Question
The equation of the tangent to the circle $x^{2}+y^{2}=16$ such that the corresponding normal passes through $(2,4)$ is(a) $2 x-y \pm 4 \sqrt{5}=0$(b) $x+2 y \pm 4 \sqrt{5}=0$(c) $\quad x-2 y \pm 4 \sqrt{5}=0$(d) $2 x+y \pm 4 \sqrt{5}=0$
Step 1
The center of the circle is at the origin (0,0). The normal to the tangent of the circle passes through the point (2,4) and the center of the circle. Show more…
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