EAPCET Mathematics · Intermediate second year · Paper IIB

Circle

Circle covers the equation of a circle and the lines related to it: tangents, normals, chords of contact, the polar of a point and the length of a tangent.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIB
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • x2+y2+2gx+2fy+c=0x^{2} + y^{2} + 2gx + 2fy + c = 0 has centre (−g,−f)(-g, -f) and radius g2+f2−c\sqrt{g^{2} + f^{2} - c}
  • Tangent at (x1,y1)(x_{1}, y_{1}): xx1+yy1+g(x+x1)+f(y+y1)+c=0xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0, written S1=0S_{1} = 0
  • Length of the tangent from (x1,y1)(x_{1}, y_{1}) =S11= \sqrt{S_{11}}
  • y=mx+cy = mx + c touches x2+y2=a2x^{2} + y^{2} = a^{2} iff c2=a2(1+m2)c^{2} = a^{2}(1 + m^{2})
  • Chord with midpoint (x1,y1)(x_{1}, y_{1}): S1=S11S_{1} = S_{11}
  • Power of a point PP with respect to a circle of centre CC and radius rr: S11=CP2−r2S_{11} = CP^{2} - r^{2} (positive outside, zero on, negative inside)
  • Circle with centre (h,k)(h, k) touching the xx-axis: (x−h)2+(y−k)2=k2(x - h)^{2} + (y - k)^{2} = k^{2}; touching the yy-axis: (x−h)2+(y−k)2=h2(x - h)^{2} + (y - k)^{2} = h^{2}
  • Circle of radius aa touching both axes (first quadrant): x2+y2−2ax−2ay+a2=0x^{2} + y^{2} - 2ax - 2ay + a^{2} = 0
  • Circle through the origin making intercepts aa and bb on the xx- and yy-axes: x2+y2−ax−by=0x^{2} + y^{2} - ax - by = 0
  • Point of contact of the tangent y=mx+cy = mx + c with x2+y2=a2x^{2} + y^{2} = a^{2}: (−ma2c,a2c)\left(-\dfrac{ma^{2}}{c}, \dfrac{a^{2}}{c}\right)
  • Pair of tangents from (x1,y1)(x_{1}, y_{1}): S12=S S11S_{1}^{2} = S\,S_{11}; chord of contact from (x1,y1)(x_{1}, y_{1}): S1=0S_{1} = 0
  • Area of the triangle formed by the two tangents from PP and their chord of contact: rL3r2+L2\dfrac{rL^{3}}{r^{2} + L^{2}}, where L=S11L = \sqrt{S_{11}}
  • Family of circles through the meet of circle S=0S = 0 and line L=0L = 0: S+λL=0S + \lambda L = 0; through the meet of two circles: S+λS′=0S + \lambda S' = 0

What this chapter covers

  • Standard and general equations
  • Position of a point with respect to a circle
  • Tangents and normals
  • Length of a tangent and power of a point
  • Chord of contact, pole and polar
  • Chord with a given midpoint

How it is usually asked

  • Find a centre and radius
  • Write a tangent or normal
  • Find the length of a tangent
  • Find a chord with a given midpoint

Understanding Circle

A circle is the set of points at a fixed distance r from a fixed centre (h, k), so its equation is (x − h)² + (y − k)² = r². Expanding gives the general form x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c). An equation of this shape is a real circle only if g² + f² − c ≥ 0, and the coefficients of x² and y² must be equal with no xy-term.

The S notation keeps the rest of the chapter compact. Write S for the left side of the circle’s equation, S₁₁ for its value at a point (x₁, y₁), and S₁ for the expression xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c. The sign of S₁₁ says whether the point is inside, on or outside the circle, and √S₁₁ is the length of the tangent from an outside point.

The same notation gives the main lines: S₁ = 0 is the tangent at a point on the circle, the chord of contact from an outside point and the polar of any point, while S₁ = S₁₁ is the chord with midpoint (x₁, y₁). A line y = mx + c touches x² + y² = a² when c² = a²(1 + m²). EAPCET questions ask for centres, radii, tangents, tangent lengths and chords.

Worked example

EAPCET-style question

The length of the tangent from the point (5,4)(5, 4) to the circle x2+y2−2x−4y−4=0x^{2} + y^{2} - 2x - 4y - 4 = 0 is

  1. (1)11\sqrt{11}
  2. (2)11
  3. (3)7\sqrt{7}
  4. (4)3

Solution

  1. S11=52+42−2(5)−4(4)−4=25+16−10−16−4=11S_{11} = 5^{2} + 4^{2} - 2(5) - 4(4) - 4 = 25 + 16 - 10 - 16 - 4 = 11.
  2. S11>0S_{11} > 0, so the point is outside the circle.
  3. Length of the tangent =S11=11= \sqrt{S_{11}} = \sqrt{11}.
  4. Check: the centre is (1,2)(1, 2) with radius 3, the distance from (5,4)(5, 4) is 20\sqrt{20}, and 20−9=11\sqrt{20 - 9} = \sqrt{11}.

Answer: 11\sqrt{11}

Shortcuts and results to remember

  • If a line through an external point PP cuts the circle at AA and BB, the tangent length PTPT satisfies PT2=PA⋅PBPT^{2} = PA \cdot PB.
  • The smallest circle through two given points is the circle on the segment joining them as diameter.
  • When only an end point and a distance along a line are known, use the parametric form (x1+rcos⁡θ,y1+rsin⁡θ)(x_{1} + r\cos\theta, y_{1} + r\sin\theta) to reach the other point.

More practice questions

Practice question 1

Tangents are drawn from (4,3)(4, 3) to the circle x2+y2=9x^{2} + y^{2} = 9. The area of the triangle formed by the tangents and their chord of contact is

  1. (1)19225\dfrac{192}{25}
  2. (2)4825\dfrac{48}{25}
  3. (3)645\dfrac{64}{5}
  4. (4)12

Solution

  1. Tangent length L=S11=16+9−9=4L = \sqrt{S_{11}} = \sqrt{16 + 9 - 9} = 4, radius r=3r = 3
  2. Area =rL3r2+L2= \dfrac{rL^{3}}{r^{2} + L^{2}}
  3. =3×649+16=19225= \dfrac{3 \times 64}{9 + 16} = \dfrac{192}{25}

Answer: 19225\dfrac{192}{25}

Practice question 2

The circle through the origin that cuts off intercepts 4 and 6 on the positive xx- and yy-axes is

  1. (1)x2+y2−4x−6y=0x^{2} + y^{2} - 4x - 6y = 0
  2. (2)x2+y2+4x+6y=0x^{2} + y^{2} + 4x + 6y = 0
  3. (3)x2+y2−2x−3y=0x^{2} + y^{2} - 2x - 3y = 0
  4. (4)x2+y2−6x−4y=0x^{2} + y^{2} - 6x - 4y = 0

Solution

  1. The circle passes through (0,0)(0, 0), (4,0)(4, 0) and (0,6)(0, 6)
  2. The angle at the origin is 90∘90^\circ, so (4,0)(4, 0) and (0,6)(0, 6) are ends of a diameter
  3. Diameter form: (x−4)(x−0)+(y−0)(y−6)=0(x - 4)(x - 0) + (y - 0)(y - 6) = 0
  4. x2+y2−4x−6y=0x^{2} + y^{2} - 4x - 6y = 0, matching x2+y2−ax−by=0x^{2} + y^{2} - ax - by = 0

Answer: x2+y2−4x−6y=0x^{2} + y^{2} - 4x - 6y = 0

Common mistakes to avoid

  • Taking the centre of x² + y² + 2gx + 2fy + c = 0 as (g, f); it is (−g, −f).
  • Reading g, f and c before making the coefficients of x² and y² equal to 1; divide through first.
  • Giving S₁₁ itself as the length of the tangent; the length is √S₁₁.
  • Taking only c = a√(1 + m²) for tangents of slope m to x² + y² = a²; c = ±a√(1 + m²), so there are two parallel tangents.

How to prepare it

Get fluent with the S, S₁, S₁₁ notation — it carries straight into System of Circles and the conics.

Once the formulas feel familiar, test them the way the exam does: practise Circle questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Circle chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Circle part of the EAPCET Mathematics syllabus?
Yes. Circle is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Circle for EAPCET?
Get fluent with the S, S₁, S₁₁ notation — it carries straight into System of Circles and the conics.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Circle?
Taking the centre of x² + y² + 2gx + 2fy + c = 0 as (g, f); it is (−g, −f). Reading g, f and c before making the coefficients of x² and y² equal to 1; divide through first. Giving S₁₁ itself as the length of the tangent; the length is √S₁₁. Taking only c = a√(1 + m²) for tangents of slope m to x² + y² = a²; c = ±a√(1 + m²), so there are two parallel tangents.