If the tangent at (1,7) to the curve x^2=y-6 | vedclass

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(B) The given curve is $x^2 = y - 6$,which can be written as $y = x^2 + 6$.To find the slope of the tangent at $(1, 7)$,we differentiate $y$ with resp

Vedclass · Accepted Answer

$95$

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(B) The given curve is $x^2 = y - 6$,which can be written as $y = x^2 + 6$.To find the slope of the tangent at $(1, 7)$,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x$.At the point $(1, 7)$,the slope $m = 2(1) = 2$.The equation of the tangent line at $(1, 7)$ is $y - 7 = 2(x - 1)$,which simplifies to $2x - y + 5 = 0$.This line touches the circle $x^2 + y^2 + 16x + 12y + C = 0$. The center of the circle is $(-8, -6)$ and the radius $r = \sqrt{(-8)^2 + (-6)^2 - C} = \sqrt{64 + 36 - C} = \sqrt{100 - C}$.The perpendicular distance from the center $(-8, -6)$ to the line $2x - y + 5 = 0$ must be equal to the radius $r$:$r = \frac{|2(-8) - (-6) + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|-16 + 6 + 5|}{\sqrt{5}} = \frac{|-5|}{\sqrt{5}} = \sqrt{5}$.Squaring both sides,we get $r^2 = 5$,so $100 - C = 5$,which implies $C = 95$.

If the tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+C=0$,then $C=$

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