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10-1.Circle and System of Circles
normal
If a circle passes through the point $(a , b) \&$ cuts the circle $x^2 + y^2= K^2$ orthogonally, then the equation of the locus of its centre is :
A
$2ax + 2by - (a^2 + b^2 + K^2) = 0$
B
$2ax + 2by - (a^2 - b^2+ K^2) = 0$
C
$x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - K^2) = 0$
D
$x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - K^2) = 0$
Solution
Let the circle be $x^{2}+y^{2}+2 g x+2 f y+c=0$
since it cuts $x^{2}+y^{2}-k^{2}=0$ orthogonally therefore
$2 g(0)+2 f(0)=c-k^{2}=0 \therefore c=k^{2}$
Again the circle $C_{2}$ passes through $(a, b)$ $\therefore a^{2}+b^{2}+2 g a+2 f b+k^{2}=0 \because c=k^{2}$
$\therefore$ Locus of centre $(-g,-f)$ is $a^{2}+b^{2}+2 a(-x)+2 b(-y)+k^{2}=0$
or $2 a x+2 b y-\left(a^{2}+b^{2}+k^{2}\right)=0$
Standard 11
Mathematics
