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  4. A circle x2+y2-6 x-16=0 cuts the x-axis at A and B and positive y-axis at the point D(0, d). The value of d equals

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Q. A circle $x^{2}+y^{2}-6 x-16=0$ cuts the $x$-axis at $A$ and $B$ and positive $y$-axis at the point $D(0, d)$. The value of $d$ equals

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Solution:

$\frac{d}{2} \cdot \frac{d-0}{0-8}=-1$
$d^{2}=16$
image
$\Rightarrow d = 4$