Circle Ques 32
- If a circle passes through the point $(a, b)$ and cuts the circle $x^{2}+y^{2}=k^{2}$ orthogonally, then the equation of the locus of its centre is
$(1988,2 M)$
(a) $2 a x+2 b y-\left(a^{2}+b^{2}+k^{2}\right)=0$
(b) $2 a x+2 b y-\left(a^{2}-b^{2}+k^{2}\right)=0$
(c) $x^{2}+y^{2}-3 a x-4 b y+a^{2}+b^{2}-k^{2}=0$
(d) $x^{2}+y^{2}-2 a x-3 b y+\left(a^{2}-b^{2}-k^{2}\right)=0$
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Answer:
Correct Answer: 32.(a)
Solution:
Formula:
- Let $x^{2}+y^{2}+2 g x+2 f y+c=0$, cuts $x^{2}+y^{2}=k^{2}$ orthogonally.
$ \begin{aligned} \Rightarrow & 2 g _1 g _2+2 f _1 f _2 & =c _1+c _2 \\ \Rightarrow & -2 g \cdot 0-2 f \cdot 0 & =c-k^{2} \\ \Rightarrow & c & =k^{2} ……(i) \end{aligned} $
Also, $x^{2}+y^{2}+2 g x+2 f y+k^{2}=0$ passes through $(a, b)$.
$\therefore \quad a^{2}+b^{2}+2 g a+2 f b+k^{2}=0 ……(ii)$
$\Rightarrow$ Required equation of locus of centre is
or
$ \begin{aligned} -2 a x-2 b y+a^{2}+b^{2}+k^{2} & =0 \\ 2 a x+2 b y-\left(a^{2}+b^{2}+k^{2}\right) & =0 \end{aligned} $
BETA
