QuizSolver
  • Bank PO
  • CBSE
  • IIT JEE
 
 

Solved Subjective Questions on Circle Set 8

Posted on - 24-02-2017

JEE Math Circle

IIT JEE

Example 1.

A circle of diameter 13 m with the centre O coinciding with the origin of co-ordinate axes, has the diameter AB on the x-axis (x co-ordinates of B > 0). If the length of the chord AC be 5m, find the equation of pair of lines BC, C having two possible positions. .

Solution

Given AB = 13m, AC = 5m. Let OL ^AC, then L is the mid-point of AC. .

Now AL =and AO =

From D ALO, cosq =

\ slope of BC =tan(90° + q) = – cotq =

Again slope of BC¢

= tan(– 90° – q) = cotq =

\ Equation of BC and BC¢ are

or y

and or

\their joint equation is y2 -

or

or 576y2 –100x2 +1300x –4225 = 0

100x2 – 576y2 –1300x + 4225 = 0 .

Example 2.

If S1 º x2 + y2 +2g1x + 2f1y +c1 = 0 and S2 º x2 + y2 +2g2x + 2f2y +c2 = 0 are two circles with radii r1 and r2 respectively, show that the points at which the circles subtend equal angles lie on the circle .

Solution

If the angle subtended by the circles S1 = 0 and S­2 = 0 at point A are q then
∠SAR = ∠PAQ = q

\ ∠C1AR = ∠PAC2 =

Let A(h, k). Then AR = , AP = .


&⇒

&⇒

\ locus of (h, k) is i.e. .

Example 3.

A chord of the circle x2 + y2 - ax - by = 0, drawn from the point (a, b), is divided by the x-axis in the ratio 2 : 1. Prove that a2 > 3b2. .

Solution

The given point A(a, b) lies on the given circle. Any point on the circle is P. The x-axis cuts the chord AP in the ratio 2:1. Hence = 0
&⇒
sinq = -


&⇒
|sinq| = < 1
&⇒
4b2 < a2 + b2
&⇒
a2 > 3b2 .

Example 4.

Find the equation to the straight lines joining the origin to the points in which the straight line y = mx +c cuts the circle x2 + y2 =2ax + 2by. .

Hence find the condition that these points may subtend a right angle at the origin. Find also the condition that the straight line may touch the circle.

Solution

The straight line is given as y =mx + c or ….(1)

The equation of the circle is given as

x2 + y2 = 2ax + 2by or x2 + y2 –2ax – 2by =0 ….(2).

Making the second equation homogeneous with the help of (1), we get the combined equation of the straight lines joining the points of intersection of (1) with (2) to the origin, hence the required equation is

x2 + y2 –2 ax= 0

or cx2 + cy2 –2ax(y – mx) – 2by(y –mx) = 0

or x2 ( c+2am) + y2 (c –2b) –2xy(a –bm) = 0 . . . (3).

If the lines represented by this equation are at right angles, the sum of the coefficients of x2 and y2 should be zero. Hence the required condition is
c + 2am + c –2b = 0 or c = b – am.

If the line (1) touches the circle, it isonly possible when the line cuts the circle in two coincident points. Therefore the lines joining the origin to the two points will also coincide. Hence the equation (3) must represent one line only. It is only possible when it is a perfect square, so its discriminant must be zero. .

So [ 2 (a – bm)]2 –4(c +2am) (c – 2b) = 0

Or a2 + b2m2 – 2abm –c2 +2bc –2amc + 4abm = 0

Or c2 – 2c (b – am) –a2 - b2m2 – 2abm = 0

Or c2 – 2c (b –am) – (a + bm)2 =0

\ c =

= b – am ± Ö [ ( b2 + a2 ) ( 1+m2 )] . .

Example 5.

DABC has vertices A(a cosa, a sina), B( a cosb, a sinb) and C(a cosg, a sing). Prove that its orthocentre is given by .

H{a(cosa + cosb + cosg), a(sina + sinb + sing)}. .

Solution

The DABC is inscribed in circle x2 + y2 = a2

so circumcentre is origin O

centroid is

Centroid divide line going circumcentre and orthocentre (h, k) in ratio 1 : 2. .

\


&⇒
h = a(cosa + cosb + cosg)

Similarly k = a(sina + sinb + sing).

 
Quadratic Equations - Solved Objective Questions Part 2 for Conceptual Clarity
Quadratic Equations - Solved Objective Questions Part 1 for Conceptual Clarity
Solved Objective Question on Probability Set 2
Solved Objective Question on Probability Set 1
Solved Objective Question on Progression and Series Set 2
Solved Objective Question on Permutations and Combinations Set 3
Solved Objective Question on Permutations and Combinations Set 2
Solved Objective Question on Progression and Series Set 1
Solved Objective Question on Permutations and Combinations Set 1
Quadratic Equations - Solved Subjective Questions Part 4
Quadratic Equations - Solved Subjective Questions Part 2
Quadratic Equations - Solved Subjective Questions Part 3
Quadratic Equations - Solved Subjective Questions Part 1
Solved Subjective Questions on Circle Set 9
Solving Equations Reducible to Quadratic Equations
Theory of Polynomial Equations and Remainder Theorem
Solving Quadratic Inequalities Using Wavy Curve Methods
Division and Distribution of Objects - Permutation and Combination
Basics of Quadratic Inequality or Inequations
Basic Concepts Of Combinations for IIT and Other Engineering Exams

Comments