Chapter 10 — CONIC SECTIONS

Theory

A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point (called the centre) remains constant (called the radius).

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Solution Roadmap

• Identify centre \((h, k)\) and radius \(r\)
• Substitute values into standard equation
• Simplify if required to obtain final form

Solution

Given centre = \((0, 2)\), radius = \(2\)

Comparing with standard form: \[ h = 0,\quad k = 2,\quad r = 2 \]

Substituting into the equation: \[ (x - 0)^2 + (y - 2)^2 = 2^2 \]

\[ x^2 + (y - 2)^2 = 4 \]

Expanding: \[ x^2 + y^2 - 4y + 4 = 4 \]

\[ x^2 + y^2 - 4y = 0 \]

Required equation of the circle: \[ x^2 + y^2 - 4y = 0 \]

Illustration

Significance for Exams

• Direct formula-based question frequently asked in CBSE board exams
• Foundation concept for advanced problems involving shifting of origin
• Important for competitive exams like JEE, NEET where circle transformations are used
• Helps in understanding general vs standard form conversion

Theory

A circle is the set of all points in a plane that are at a constant distance from a fixed point called the centre.

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Solution Roadmap

• Identify centre \((h, k)\) and radius \(r\)
• Substitute values into standard equation
• Simplify to obtain required form

Solution

Given centre = \((-2, 3)\), radius = \(4\)

Comparing with standard form: \[ h = -2,\quad k = 3,\quad r = 4 \]

Substituting into the equation: \[ (x - (-2))^2 + (y - 3)^2 = 4^2 \]

\[ (x + 2)^2 + (y - 3)^2 = 16 \]

Expanding: \[ x^2 + 4x + 4 + y^2 - 6y + 9 = 16 \]

\[ x^2 + y^2 + 4x - 6y - 3 = 0 \]

Required equation: \[ x^2 + y^2 + 4x - 6y - 3 = 0 \]

Illustration

Significance for Exams

• Standard direct substitution question frequently asked in board exams
• Builds base for completing square method in reverse problems
• Useful in coordinate geometry problems involving locus
• Common in JEE/NEET for quick equation formation

Theory

The equation of a circle with centre \((h, k)\) and radius \(r\) is derived using the distance formula:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

This represents all points whose distance from the centre remains constant.

Solution Roadmap

• Substitute fractional centre and radius into standard form
• Expand carefully (handle fractions systematically)
• Combine constants correctly
• Multiply to eliminate fractions for final form

Solution

Given centre = \(\left(\dfrac{1}{2}, \dfrac{1}{4}\right)\), radius = \(\dfrac{1}{12}\)

Comparing with standard form: \[ h = \dfrac{1}{2},\quad k = \dfrac{1}{4},\quad r = \dfrac{1}{12} \]

Substituting: \[ \left(x - \dfrac{1}{2}\right)^2 + \left(y - \dfrac{1}{4}\right)^2 = \left(\dfrac{1}{12}\right)^2 \]

Expanding: \[ x^2 - x + \dfrac{1}{4} + y^2 - \dfrac{y}{2} + \dfrac{1}{16} = \dfrac{1}{144} \]

Combine constants: \[ \dfrac{1}{4} + \dfrac{1}{16} = \dfrac{5}{16} \]

\[ x^2 + y^2 - x - \dfrac{y}{2} + \dfrac{5}{16} = \dfrac{1}{144} \]

Bringing all terms to one side: \[ x^2 + y^2 - x - \dfrac{y}{2} + \left(\dfrac{5}{16} - \dfrac{1}{144}\right) = 0 \]

\[ \dfrac{5}{16} = \dfrac{45}{144} \Rightarrow \dfrac{45}{144} - \dfrac{1}{144} = \dfrac{44}{144} = \dfrac{11}{36} \]

\[ x^2 + y^2 - x - \dfrac{y}{2} + \dfrac{11}{36} = 0 \]

Multiplying throughout by \(36\): \[ 36x^2 + 36y^2 - 36x - 18y + 11 = 0 \]

Required equation: \[ 36x^2 + 36y^2 - 36x - 18y + 11 = 0 \]

Illustration

Significance for Exams

• Tests algebraic accuracy with fractions (very common in CBSE exams)
• Important for mastering expansion and simplification techniques
• Helps in handling rational coordinates in coordinate geometry
• Frequently appears in JEE/NEET as part of multi-step problems

Theory

A circle represents all points in a plane that are at a fixed distance from a given point called the centre.

The standard equation of a circle is: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Solution Roadmap

• Identify centre \((h, k)\) and radius \(r\)
• Substitute into standard equation
• Expand and simplify carefully

Solution

Given centre = \((1, 1)\), radius = \(\sqrt{2}\)

Comparing: \[ h = 1,\quad k = 1,\quad r = \sqrt{2} \]

Substituting: \[ (x - 1)^2 + (y - 1)^2 = (\sqrt{2})^2 \]

\[ (x - 1)^2 + (y - 1)^2 = 2 \]

Expanding: \[ x^2 - 2x + 1 + y^2 - 2y + 1 = 2 \]

\[ x^2 + y^2 - 2x - 2y + 2 = 2 \]

\[ x^2 + y^2 - 2x - 2y = 0 \]

Required equation: \[ x^2 + y^2 - 2x - 2y = 0 \]

Illustration

Significance for Exams

• Tests understanding of squaring irrational radius values
• Frequently asked in board exams in direct or modified form
• Important for recognizing simplified general form quickly
• Useful in coordinate geometry problems involving translation of axes

Theory

The equation of a circle with centre \((h, k)\) and radius \(r\) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

This follows from the distance formula, ensuring every point on the circle is at a constant distance from the centre.

Solution Roadmap

• Identify symbolic centre and radius
• Substitute carefully (sign handling is critical)
• Expand and simplify algebraically

Solution

Given centre = \((-a, -b)\), radius = \(\sqrt{a^2 - b^2}\)

Comparing: \[ h = -a,\quad k = -b,\quad r = \sqrt{a^2 - b^2} \]

Substituting: \[ (x - (-a))^2 + (y - (-b))^2 = (\sqrt{a^2 - b^2})^2 \]

\[ (x + a)^2 + (y + b)^2 = a^2 - b^2 \]

Expanding: \[ x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 - b^2 \]

Bringing all terms to one side: \[ x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 = 0 \]

\[ x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \]

Required equation: \[ x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \]

Illustration

Significance for Exams

• Tests symbolic manipulation and sign handling (very common mistake area)
• Important for deriving general equation from parameters
• Useful in reverse problems (finding centre and radius from equation)
• Frequently used in JEE for parameter-based geometry questions

Important Note

For the radius to be real, the condition \(a^2 \ge b^2\) must hold.

Theory

The standard form of a circle is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

By comparing a given equation with this form, the centre \((h, k)\) and radius \(r\) can be identified directly.

Solution Roadmap

• Rewrite terms to match \((x - h)\) and \((y - k)\)
• Identify centre coordinates carefully (sign reversal)
• Take square root of constant term for radius

Solution

Given equation: \[ (x + 5)^2 + (y - 3)^2 = 36 \]

Compare with: \[ (x - h)^2 + (y - k)^2 = r^2 \]

\[ (x + 5)^2 = (x - (-5))^2 \Rightarrow h = -5 \]

\[ (y - 3)^2 \Rightarrow k = 3 \]

\[ r^2 = 36 \Rightarrow r = \sqrt{36} = 6 \]

Centre = \((-5, 3)\), Radius = \(6\)

Illustration

Significance for Exams

• Direct comparison type question frequently asked in board exams
• Tests sign interpretation: \((x + a) \Rightarrow h = -a\)
• Forms the basis for converting general equation to standard form
• Important for quick identification in competitive exams

Common Mistake

Students often take centre as \((5, 3)\) instead of \((-5, 3)\). Always reverse the sign while comparing.

Theory

The general equation of a circle is:

\[ x^2 + y^2 + Dx + Ey + F = 0 \]

To find the centre and radius, convert it into standard form by completing the square:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Solution Roadmap

• Group \(x\) and \(y\) terms
• Complete squares by adding appropriate constants
• Rewrite in standard form
• Identify centre and radius

Solution

Given: \[ x^2 + y^2 - 4x - 8y - 45 = 0 \]

Group terms: \[ x^2 - 4x + y^2 - 8y = 45 \]

Complete squares:

\[ x^2 - 4x + 4 + y^2 - 8y + 16 = 45 + 4 + 16 \]

\[ (x - 2)^2 + (y - 4)^2 = 65 \]

Comparing with standard form: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Centre = \((2, 4)\), Radius = \(\sqrt{65}\)

Illustration

Significance for Exams

• Classic completing-square problem (very frequent in CBSE boards)
• Core skill for converting general form to standard form
• Essential for JEE/NEET coordinate geometry problems
• Helps in quick identification of centre using shortcut: \((-D/2, -E/2)\)

Shortcut Insight

For equation \(x^2 + y^2 + Dx + Ey + F = 0\),

Centre = \(\left(-\dfrac{D}{2}, -\dfrac{E}{2}\right)\), Radius = \(\sqrt{\dfrac{D^2 + E^2}{4} - F}\)

Common Mistake

Students often forget to add the same constants on both sides while completing squares.

Theory

The general form of a circle is:

\[ x^2 + y^2 + Dx + Ey + F = 0 \]

To find the centre and radius, convert it into standard form using completing the square:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Solution Roadmap

• Shift constant term to RHS
• Group \(x\) and \(y\) terms
• Complete squares by adding appropriate values
• Rewrite in standard form and identify parameters

Solution

Given: \[ x^2 + y^2 - 8x + 10y - 12 = 0 \]

Rearranging: \[ x^2 - 8x + y^2 + 10y = 12 \]

Complete squares:

\[ x^2 - 8x + 16 + y^2 + 10y + 25 = 12 + 16 + 25 \]

\[ (x - 4)^2 + (y + 5)^2 = 53 \]

Comparing with: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Centre = \((4, -5)\), Radius = \(\sqrt{53}\)

Illustration

Significance for Exams

• Standard completing-square problem frequently asked in exams
• Tests correct handling of positive and negative linear terms
• Important for quick identification using shortcut formulas
• Used in advanced problems involving tangents and loci

Shortcut Insight

For equation \(x^2 + y^2 + Dx + Ey + F = 0\),

Centre = \(\left(-\dfrac{D}{2}, -\dfrac{E}{2}\right)\), Radius = \(\sqrt{\dfrac{D^2 + E^2}{4} - F}\)

Common Mistake

Students often write centre as \((4, 5)\) instead of \((4, -5)\). Always observe the sign inside the bracket carefully.

Theory

The general equation of a circle is:

\[ x^2 + y^2 + Dx + Ey + F = 0 \]

To convert it into standard form, coefficients of \(x^2\) and \(y^2\) must be equal and unity, followed by completing the square.

Solution Roadmap

• Make coefficients of \(x^2\) and \(y^2\) equal to 1
• Identify which variable needs completing square
• Convert into standard form
• Extract centre and radius

Solution

Given: \[ 2x^2 + 2y^2 - x = 0 \]

Divide throughout by \(2\): \[ x^2 + y^2 - \dfrac{x}{2} = 0 \]

Rearranging: \[ x^2 - \dfrac{x}{2} + y^2 = 0 \]

Complete square for \(x\):

\[ x^2 - \dfrac{x}{2} + \left(\dfrac{1}{4}\right)^2 = \dfrac{1}{16} \]

\[ \left(x - \dfrac{1}{4}\right)^2 + y^2 = \dfrac{1}{16} \]

Comparing with: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Centre = \(\left(\dfrac{1}{4}, 0\right)\), Radius = \(\dfrac{1}{4}\)

Illustration

Significance for Exams

• Tests normalization step (dividing equation first)
• Important case where only one variable requires completing square
• Common in board exams as a conceptual check question
• Useful in JEE for identifying shifted circles quickly

Shortcut Insight

After dividing by 2, compare with \(x^2 + y^2 + Dx + Ey + F = 0\):

\(D = -\dfrac{1}{2},\; E = 0\)

Centre = \(\left(-\dfrac{D}{2}, -\dfrac{E}{2}\right) = \left(\dfrac{1}{4}, 0\right)\)

Common Mistake

Students often forget to divide the entire equation first, leading to incorrect centre and radius.

Theory

If a circle passes through two points, then its centre lies on the perpendicular bisector of the line joining those points.

Also, the centre \((h, k)\) satisfies all given geometric conditions, and the radius is the distance from the centre to any point on the circle.

Solution Roadmap

• Assume centre \((h, k)\)
• Use equal distance condition from two given points
• Use given line condition
• Solve system to find centre
• Compute radius and form equation

Solution

Let the centre be \((h, k)\). Since the circle passes through \((4,1)\) and \((6,5)\),

\[ (4 - h)^2 + (1 - k)^2 = (6 - h)^2 + (5 - k)^2 \]

Expanding: \[ 16 - 8h + h^2 + 1 - 2k + k^2 = 36 - 12h + h^2 + 25 - 10k + k^2 \]

Simplifying: \[ 4h + 8k - 44 = 0 \Rightarrow h + 2k = 11 \]

Given condition: \[ 4h + k = 16 \]

Solving: \[ k = 16 - 4h \]

Substitute: \[ h + 2(16 - 4h) = 11 \]

\[ h + 32 - 8h = 11 \Rightarrow -7h = -21 \Rightarrow h = 3 \]

\[ k = 16 - 4(3) = 4 \]

Centre = \((3, 4)\)

Radius: \[ r^2 = (4 - 3)^2 + (1 - 4)^2 = 1 + 9 = 10 \]

Equation: \[ (x - 3)^2 + (y - 4)^2 = 10 \]

Final Answer: \[ x^2 + y^2 - 6x - 8y + 15 = 0 \]

Illustration

Significance for Exams

• Combines locus concept with circle equation (high-weight problem type)
• Frequently appears in CBSE board exams as a multi-step question
• Important for JEE/NEET: intersection of geometry + algebra
• Strengthens system-solving and geometric interpretation skills

Shortcut Insight

The centre lies on the perpendicular bisector of the line joining \((4,1)\) and \((6,5)\). This gives the same equation \(h + 2k = 11\) directly, reducing algebra.

Common Mistake

Students often forget that both points give equal radius, leading to incorrect equations. Always equate distances to eliminate \(r^2\).

Theory

If a circle passes through two points, its centre lies on the perpendicular bisector of the line joining those points.

The centre must also satisfy any additional given condition (here, a line), so we solve a system of equations to determine it.

Solution Roadmap

• Assume centre \((h, k)\)
• Use equal distance condition from given points
• Apply line condition
• Solve system to find centre
• Compute radius and form equation

Solution

Let the centre be \((h, k)\). Since the circle passes through \((2,3)\) and \((-1,1)\),

\[ (2 - h)^2 + (3 - k)^2 = (-1 - h)^2 + (1 - k)^2 \]

Expanding: \[ 4 - 4h + h^2 + 9 - 6k + k^2 = 1 + 2h + h^2 + 1 - 2k + k^2 \]

Simplifying: \[ \begin{aligned} -6h - 4k + 11 &= 0 \\\Rightarrow 6h + 4k &= 11 \end{aligned} \]

Given condition: \[ \begin{aligned} h - 3k - 11 &= 0 \\\Rightarrow h &= 3k + 11 \end{aligned} \]

Substitute: \[ 6(3k + 11) + 4k = 11 \]

\[ \begin{aligned} 18k + 66 + 4k &= 11 \\\Rightarrow 22k &= -55 \\\Rightarrow k &= -\dfrac{5}{2}\\\\ h &= 3\left(-\dfrac{5}{2}\right) + 11 \\&= \dfrac{7}{2} \end{aligned} \]

Centre = \(\left(\dfrac{7}{2}, -\dfrac{5}{2}\right)\)

Radius: \[ \begin{aligned} r^2 &= \left(\dfrac{7}{2} - 2\right)^2 + \left(-\dfrac{5}{2} - 3\right)^2\\ &= \dfrac{9}{4} + \dfrac{121}{4} = \dfrac{130}{4} = \dfrac{65}{2} \end{aligned} \]

Equation: \[ \left(x - \dfrac{7}{2}\right)^2 + \left(y + \dfrac{5}{2}\right)^2 = \dfrac{65}{2} \]

Converting to general form: \[ x^2 + y^2 - 7x + 5y - 14 = 0 \]

Final Answer: \[ x^2 + y^2 - 7x + 5y - 14 = 0 \]

Significance for Exams

• High-level problem combining locus, algebra, and geometry
• Very common in CBSE board exams (long-answer type)
• Important for JEE/NEET multi-condition problems
• Strengthens handling of fractional centres and radii

Shortcut Insight

Instead of full expansion, the perpendicular bisector of the points \((2,3)\) and \((-1,1)\) directly gives the equation \(6h + 4k = 11\), saving time.

Common Mistake

Students often make sign errors while expanding or mishandle fractions in the final steps. Careful algebra is essential.

Theory

If the centre of a circle lies on the \(x\)-axis, its coordinates are of the form \((h, 0)\).

A point lies on the circle if it satisfies the equation: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Solution Roadmap

• Assume centre \((h, 0)\)
• Use given point to form equation
• Solve for \(h\)
• Form equation(s) of circle

Solution

Let the centre be \((h, 0)\), radius = \(5\)

Equation: \[ (x - h)^2 + y^2 = 25 \]

Since the circle passes through \((2,3)\),

\[ \begin{aligned} (2 - h)^2 + 3^2 &= 25\\ (2 - h)^2 + 9 &= 25 \\\Rightarrow (2 - h)^2 &= 16\\ 2 - h &= \pm 4 \end{aligned} \]

Case 1: \(2 - h = 4 \Rightarrow h = -2\)

Case 2: \(2 - h = -4 \Rightarrow h = 6\)

Thus, two possible centres exist due to symmetry about the given point.

Equations:

\[ (x + 2)^2 + y^2 = 25 \]

\[ (x - 6)^2 + y^2 = 25 \]

Final Answer: \[ \begin{aligned} (x + 2)^2 + y^2 &= 25 \\ \text{and} \\ (x - 6)^2 + y^2 &= 25 \end{aligned} \]

Illustration

Significance for Exams

• Tests understanding of geometric symmetry and multiple solutions
• Common CBSE board question type (conceptual)
• Important for JEE where multiple loci solutions occur
• Strengthens interpretation of parameter-based centres

Key Insight

The point \((2,3)\) lies at equal distance from two points on the \(x\)-axis, leading to two possible circles.

Common Mistake

Students often give only one equation, missing the second valid solution.

Theory

The general equation of a circle is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

If a circle passes through a point, that point satisfies the equation. Intercepts on axes indicate that the circle passes through specific axis points.

Solution Roadmap

• Use general equation of circle
• Apply condition of passing through origin
• Use intercept points \((a,0)\) and \((0,b)\)
• Determine coefficients and form equation

Solution

Let the circle cut the \(x\)-axis at \(A(a,0)\) and the \(y\)-axis at \(B(0,b)\).

General equation: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

Since the circle passes through \((0,0)\),

\[ c = 0 \]

So equation becomes: \[ x^2 + y^2 + 2gx + 2fy = 0 \]

Passing through \((a,0)\):

\[ \begin{aligned} a^2 + 2ga &= 0 \\\Rightarrow 2g &= -a \end{aligned} \]

Passing through \((0,b)\):

\[ \begin{aligned} b^2 + 2fb &= 0 \\\Rightarrow 2f &= -b \end{aligned} \]

Substituting: \[ x^2 + y^2 - ax - by = 0 \]

Required equation: \[ x^2 + y^2 - ax - by = 0 \]

Illustration

Significance for Exams

• Classic derivation-based question (frequently asked in CBSE exams)
• Tests understanding of general equation and parameter identification
• Important for JEE/NEET in parameterized geometry problems
• Forms a standard result often used directly in advanced problems

Key Insight

This result is a standard form: any circle passing through origin and intercepting axes at \(a\) and \(b\) has equation \(x^2 + y^2 - ax - by = 0\).

Common Mistake

Students often forget to substitute \((0,0)\) first, leading to incorrect constant term.

Theory

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

If a point lies on the circle, then the radius is the distance between the centre and that point.

Solution Roadmap

• Identify centre \((h, k)\)
• Find radius using distance formula
• Substitute into standard equation

Solution

Given centre = \((2,2)\), point = \((4,5)\)

Radius: \[ \begin{aligned} r^2 &= (4 - 2)^2 + (5 - 2)^2 \\&= 4 + 9 \\&= 13 \end{aligned} \]

Equation: \[ (x - 2)^2 + (y - 2)^2 = 13 \]

Required equation: \[ (x - 2)^2 + (y - 2)^2 = 13 \]

Illustration

Significance for Exams

• Direct application of distance formula (very common in CBSE exams)
• Tests quick identification of radius from given point
• Frequently used as a base step in complex geometry problems
• Important for speed-solving in competitive exams

Shortcut Insight

No need to find \(r\) separately. Directly compute \(r^2\) using distance formula and substitute.

Common Mistake

Students sometimes calculate \(r\) instead of \(r^2\), leading to unnecessary square root calculations.

Theory

For a circle \(x^2 + y^2 = r^2\), the centre is \((0,0)\) and radius is \(r\).

To determine the position of a point \((x_1, y_1)\), compare:

• If \(x_1^2 + y_1^2 = r^2\) → point lies on the circle
• If \(x_1^2 + y_1^2 < r^2\) → point lies inside
• If \(x_1^2 + y_1^2 > r^2\) → point lies outside

Solution Roadmap

• Identify centre and radius
• Compute \(x^2 + y^2\) for given point
• Compare with \(r^2\)

Solution

Given: \[ x^2 + y^2 = 25 \Rightarrow r^2 = 25 \]

For point \((-2.5, 3.5)\):

\[ \begin{aligned} (-2.5)^2 + (3.5)^2 &= 6.25 + 12.25 \\&= 18.5 \end{aligned} \]

Compare: \[ 18.5 < 25 \]

Therefore, the point lies inside the circle.

Illustration

Significance for Exams

• Direct concept-based question frequently asked in CBSE exams
• Important for JEE/NEET in geometry and locus problems
• Used in testing position relative to circle (inside/on/outside)
• Foundation for inequalities involving circles

Shortcut Insight

Always compare with \(r^2\), not \(r\), to avoid unnecessary square root calculations.

Common Mistake

Students often compare distance instead of squared distance, leading to extra computation or errors.