Question
Locus of point of intersection of perpendicular tangents to the circle $x^{2}+y^{2}-4 x-6 y-12=0$(a) $x^{2}+y^{2}-4 x-6 y+37=0$(b) $x^{2}+y^{2}-4 x-6 y-37=0$(c) $x^{2}+y^{2}+4 x+6 y-37=0$(d) $x^{2}+y^{2}-4 x-6 y-25=0$
Step 1
The center of the circle $(h, k)$ can be found using the formula $h = -\frac{b}{2a}$ and $k = -\frac{d}{2c}$, where $a$, $b$, $c$, and $d$ are the coefficients of the equation. So, the center of the circle is $(2, 3)$. Show more…
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