If a circle passes through the point (a, b) a | vedclass

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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it cuts the circle $x^2 + y^2 - K^2 = 0$ orthogonally,the condition $2g_1g

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

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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it cuts the circle $x^2 + y^2 - K^2 = 0$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ applies.Here,$g_2 = 0, f_2 = 0$,and $c_2 = -K^2$. Thus,$2g(0) + 2f(0) = c - K^2$,which implies $c = K^2$.Since the circle passes through the point $(a, b)$,we have $a^2 + b^2 + 2ga + 2fb + c = 0$.Substituting $c = K^2$,we get $a^2 + b^2 + 2ga + 2fb + K^2 = 0$.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 $a^2 + b^2 + 2(-x)a + 2(-y)b + K^2 = 0$.Therefore,the locus of the centre is $2ax + 2by - (a^2 + b^2 + K^2) = 0$.

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:

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