The point at which the normal to the circle x | vedclass

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(C) The equation of the circle is $x^2 + y^2 + 4x + 6y - 39 = 0$. The center of the circle is $(-g, -f) = (-2, -3)$.The normal to a circle at any poin

Vedclass · Accepted Answer

$(-6, -9)$

Vedclass · Answer

(C) The equation of the circle is $x^2 + y^2 + 4x + 6y - 39 = 0$. The center of the circle is $(-g, -f) = (-2, -3)$.The normal to a circle at any point always passes through the center of the circle.The normal passes through the point $(2, 3)$ and the center $(-2, -3)$.The slope of the normal is $m = \frac{3 - (-3)}{2 - (-2)} = \frac{6}{4} = \frac{3}{2}$.The equation of the normal line is $y - 3 = \frac{3}{2}(x - 2)$,which simplifies to $2y - 6 = 3x - 6$,or $3x - 2y = 0$.To find the other point of intersection,substitute $x = \frac{2y}{3}$ into the circle equation:$(\frac{2y}{3})^2 + y^2 + 4(\frac{2y}{3}) + 6y - 39 = 0$$\frac{4y^2}{9} + y^2 + \frac{8y}{3} + 6y - 39 = 0$Multiply by $9$: $4y^2 + 9y^2 + 24y + 54y - 351 = 0$$13y^2 + 78y - 351 = 0$Divide by $13$: $y^2 + 6y - 27 = 0$$(y - 3)(y + 9) = 0$,so $y = 3$ or $y = -9$.If $y = 3$,$x = \frac{2(3)}{3} = 2$ (this is the given point).If $y = -9$,$x = \frac{2(-9)}{3} = -6$.Thus,the other point is $(-6, -9)$.

The point at which the normal to the circle $x^2 + y^2 + 4x + 6y - 39 = 0$ at the point $(2, 3)$ meets the circle again is:

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