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

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

Vedclass · Accepted Answer

$95$

Vedclass · Answer

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

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 the value of $c$ is:

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