If P(-9,-1) is a point on the circle x^2+y^2+ | vedclass

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(B) The given circle is $x^2+y^2+4x+8y-38=0$. Completing the square,we get $(x+2)^2+(y+4)^2=58$. The center of the circle is $C(-2,-4)$. Let $Q(x_1, y

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$7x-3y=56$

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(B) The given circle is $x^2+y^2+4x+8y-38=0$. Completing the square,we get $(x+2)^2+(y+4)^2=58$. The center of the circle is $C(-2,-4)$. Let $Q(x_1, y_1)$ be the other end of the diameter passing through $P(-9,-1)$. Since $C$ is the midpoint of $PQ$,we have $\frac{x_1-9}{2}=-2 \Rightarrow x_1=5$ and $\frac{y_1-1}{2}=-4 \Rightarrow y_1=-7$. Thus,$Q$ is $(5,-7)$. The tangent at $Q$ is parallel to the tangent at $P$. The slope of the radius $CP$ is $m_{CP} = \frac{-1-(-4)}{-9-(-2)} = \frac{3}{-7} = -\frac{3}{7}$. The slope of the tangent at $P$ (and $Q$) is the negative reciprocal of the slope of the radius,which is $m = -\frac{1}{-3/7} = \frac{7}{3}$. The equation of the tangent at $Q(5,-7)$ is $y - (-7) = \frac{7}{3}(x - 5)$. $3(y+7) = 7(x-5)$ $\Rightarrow 3y+21 = 7x-35$ $\Rightarrow 7x-3y=56$.

If $P(-9,-1)$ is a point on the circle $x^2+y^2+4x+8y-38=0$,then find the equation of the tangent drawn at the other end of the diameter drawn through $P$.

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