Test Index

Circles Part 1

Section: Mathematics

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Question : 91 of 100

Marks: +1, -0

A variable circle passes through the fixed point

A(p, q)

x

-axis. The locus of the other end of the diameter through

A

$(y-q)^2=4 p x$
$(x-q)^2=4 p y$
$(y-p)^2=4 q x$
$(x-p)^2=4 q y$

Solution:

Let the variable circle be

x^2+y^2+2 g x+2 f y+c=0

........(1)

\therefore p^2+q^2+2 g p+2 f q+c=0

......(2)

Circle (1) touches

x

-axis,

\therefore g^2-c=0 \Rightarrow c=g^2

.

From (2)

p^2+q^2+2 g p+2 f q+g^2=0

.....(3)

Let the other end of diameter through

(p, q)

be

(h, k)

,

then,

\; \frac{h+p}{2} = -g

and

\; \frac{k+q}{2} = -f

Put in (3)

p^2+q^2+2 p \left(-\; \frac{h+p}{2}\right) +2 q \left(-\; \frac{k+q}{2}\right) + \left(\; \frac{h+p}{2}\right)^2 =0

\Rightarrow h^2+p^2-2 h p-4 k q=0

\therefore

locus of

(h, k)

is

x^2+p^2-2 x p-4 y q=0

\Rightarrow (x-p)^2=4 q y

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