If a circle passes through the point (1, 2) a | vedclass

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(1, 2)$,we have $1^2 + 2^2 + 2g(1) + 2f(2) + c = 0$,whi

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$2x + 4y - 9 = 0$

Vedclass · Answer

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(1, 2)$,we have $1^2 + 2^2 + 2g(1) + 2f(2) + c = 0$,which simplifies to $2g + 4f + c + 5 = 0$.Since the circle intersects $x^2 + y^2 = 4$ (or $x^2 + y^2 - 4 = 0$) orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ applies.Here,$g_1 = g, f_1 = f, c_1 = c$ and $g_2 = 0, f_2 = 0, c_2 = -4$.Thus,$2g(0) + 2f(0) = c - 4$,which gives $c = 4$.Substituting $c = 4$ into the first equation: $2g + 4f + 4 + 5 = 0$,so $2g + 4f + 9 = 0$.The center of the circle is $(-g, -f)$. Let the center be $(x, y)$,so $g = -x$ and $f = -y$.Substituting these into the equation $2g + 4f + 9 = 0$,we get $2(-x) + 4(-y) + 9 = 0$,which simplifies to $2x + 4y - 9 = 0$.

If a circle passes through the point $(1, 2)$ and intersects the circle $x^2 + y^2 = 4$ orthogonally,then the equation of the locus of its center is:

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