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If a circle passes through the point $(a, b)$ and cuts the circle $x^2+y^2=p^2$ orthogonally, then the equation of the locus of its centre is
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The correct answer is:
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$
Let the centre be $(\alpha, \beta)$
$\because$ It cut the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{p}^2$ orthogonally
$2(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2$
$\mathrm{c}_1=\mathrm{p}^2$
Let equation of circle is $x^2+y^2-2 \alpha x-2 \beta y+p^2=0$
It pass through $(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0$
Locus $\therefore 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$
$\because$ It cut the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{p}^2$ orthogonally
$2(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2$
$\mathrm{c}_1=\mathrm{p}^2$
Let equation of circle is $x^2+y^2-2 \alpha x-2 \beta y+p^2=0$
It pass through $(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0$
Locus $\therefore 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$
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