Let S = 0 be the locus of the center of a var | vedclass

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(B) Let the variable circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since it passes through $(4, 6)$,we have $16 + 36 + 8g + 12f + c = 0$,or $8g + 12f + c

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$2\sqrt{13}$

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(B) Let the variable circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since it passes through $(4, 6)$,we have $16 + 36 + 8g + 12f + c = 0$,or $8g + 12f + c = -52$. Since it intersects $x^2 + y^2 - 4x - 6y = 0$ orthogonally,$2g(-2) + 2f(-3) = c + 0$,so $c = -4g - 6f$. Substituting $c$ into the first equation: $8g + 12f - 4g - 6f = -52$,which simplifies to $4g + 6f = -52$,or $2g + 3f = -26$. The center of the variable circle is $(-g, -f)$. Let this be $(x, y)$,so $g = -x$ and $f = -y$. Substituting these into the equation $2g + 3f = -26$,we get $2(-x) + 3(-y) = -26$,which is $2x + 3y = 26$. This is the equation of the locus $S = 0$. The least distance from the origin $O(0, 0)$ to the line $2x + 3y - 26 = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$. $d = \frac{|2(0) + 3(0) - 26|}{\sqrt{2^2 + 3^2}} = \frac{26}{\sqrt{4 + 9}} = \frac{26}{\sqrt{13}} = 2\sqrt{13}$.

Let $S = 0$ be the locus of the center of a variable circle which intersects the circle $x^2 + y^2 - 4x - 6y = 0$ orthogonally at the point $(4, 6)$. If $P$ is a variable point on $S = 0$,then the least value of $OP$ is (where $O$ is the origin).

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