(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it cuts the circle $x^2 + y^2 = p^2$ orthogonally,the condition $2g_1g_2 +

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$2ax + 2by - (a^2 + b^2 + p^2) = 0$

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it cuts the circle $x^2 + y^2 = p^2$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives $2g(0) + 2f(0) = c - p^2$,which implies $c = p^2$.Since the circle passes through $(a, b)$,we have $a^2 + b^2 + 2ga + 2fb + p^2 = 0$.The centre of the circle is $(-g, -f)$. Let the centre be $(x, y)$,so $g = -x$ and $f = -y$.Substituting these into the equation,we get $a^2 + b^2 + 2(-x)a + 2(-y)b + p^2 = 0$.This simplifies to $2ax + 2by - (a^2 + b^2 + p^2) = 0$.

Class 11 Mathematics 10-1.Circle and System of Circles Locus Related Problem

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The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

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IIT 1988 All IIT

A
$2ax + 2by - (a^2 + b^2 + p^2) = 0$
B
$2ax + 2by - (a^2 - b^2 + p^2) = 0$
C
$x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - p^2) = 0$
D
$x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - p^2) = 0$

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The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

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Let $PQ$ and $MN$ be two straight lines touching the circle $x^{2}+y^{2}-4x-6y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB=\pi/3$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ is:

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