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Solved Subjective Questions on Circle Set 6

Posted on - 20-02-2017

JEE Math Circle

IIT JEE

Example 1.

Find the coordinates of the points at which the circles x2 + y2 - 4x - 2y = 4 and
x2 + y2 - 12x - 8y = 12 touch each other
. Find the coordinates of the point of contact and equation of the common tangent at the point of contact.

Solution

The given circles are

S1 º x2 + y2 – 4x – 2y – 4 = 0 and S2 º x2 + y2 – 12x – 8y – 12 = 0

Equation of the common tangent is S1 – S2 = 0 i.e. 8x + 6y + 8 = 0.


&⇒
4x + 3y + 4 = 0
. . . (1).

Let the point of contact be (x1, y1)

Equation of the tangent at (x1, y1) is ,

xx1 + yy1 – 2(x + x1) – (y + y1) – 4 = 0

or (x1 – 2)x + (y1 – 1)y – 2x1 – y1 – 4 = 0 . . . (2).

Comparing (1) and (2), we get


&⇒
x1 = 4k + 2, y1 = 3k + 1, 2x1 + y1 + 4 = – 4k


&⇒
2(4k + 2) + 3k + 1 + 4 = – 4k or 8k + 4 + 3k + 5 = – 4k


&⇒
15k = –9
&⇒
k =
&⇒
x1 = 4 ´ , y1 = 3 ´

Point of contact =

Example 2.

A variable triangle has two of its sides along the axes. Its 3rd side touches the circle x2 + y2 - 2ax - 2ay + a2 = 0. Prove that the locus of the circumcentre of the triangle is a2 - 2a(x + y) + 2xy = 0.

Solution

The given circle has its centre (a, a) and radius a, so that the circle touches both the axes. Let the side AB be
&⇒
A º (p, o), B º (o, q)

Let R be the midpoint of AB, so that

R º = (h, k)
&⇒
p = 2h, q = 2k

Since the line touches the circle,

So,
&⇒

(ap + aq – pq)2 = a2 (p2 +q2)


&⇒
a2(p+q)2+p2q2–2a(p+q)×pq
= a2(p2 + q2)


&⇒
2a2 +pq – 2a (p + q) = 0


&⇒
a2 + 2hk – 2a (h + k) = 0

Locus of (h, k) is

a2 + 2xy – 2a (x + y) = 0

Example 3.

Consider a curve ax2 + 2hxy + by2 = 1, and a point P not lying on the curve. A line drawn from the point P intersects the curve at points Q and R. If the product PQ.PR is independent of the slope of the line, show that the curve is a circle.

Solution

Let P be the point (a, b). The equation of the line through P, with slope tanq, is = r

Any point on this line is (a + r cosq, b + r sinq). If this point lies on the curve, then .

a(a + r cosq)2 + 2h(a + r cosq) (b + r sinq) + b(b + r sinq)2 = 1

This is a quadratic equation in r whose values give the coordinates of Q and R, where PQ = r1 and PR = r2.


&⇒
r1r2 =

=

which is independent of q if + h2 = 0
&⇒
a = b, h = 0.

Hence the given curve is a circle.

Example 4.

Find the equation of the circle passing through (-4, 3) and touching the lines
x + y = 2 and x - y = 2.

Solution

The lines are mutually perpendicular and passes through (2, 0). .

Due to symmetry, obviously the centre of the required circle lies on x-axis. Let it be (-a, 0).

So equation of circle is

x2 + y2 + 2ax + c = 0

It passes through (-4, 3)

\ 16 + 9 – 8a + c = 0
&⇒
8a – 25 = c

radius =

But radius = length of perpendicular from (-a, 0) on x + y = 2. .

This gives a = 10 ± 3


&⇒
c = 55 ± 24

circle is x2 + y2 + 2(10 ± 3)x + 55 ± 24= 0

Example 5.

Find the circumcentre of the triangle formed by x + y =0, x – y = 0 and lx + my =1. If l and m vary such that l2 + m2 = 1, show that the locus of its circumcentre is the curve (x2 – y2)2 = x2 + y2 .

Solution

Solving the straight lines x – y = 0, x + y = 0, lx + my = 1

we get the vertices as A(0, 0), Band

if S(h, k) is the circumcentre then SA2 = SB2 = SC2 and hence

h2 + k2 = =

or h + k = ….(1)

or h – k = ….(2)

if l, m vary such that l2 + m2 = 1, then the locus of the circumcentre is got from

h2 + k2 = = (h2 – k2)2

Hence locus of circumcentre is (x­2 – y2)2 = x2 + y2 . .

 
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