The tangent to the parabola y = x^2 + 6 at th | vedclass

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(C) The equation of the tangent to the parabola $y = x^2 + 6$ at $(1, 7)$ is given by $\frac{1}{2}(y + 7) = x(1) + 6$.Simplifying this,we get $y + 7 =

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$(-6, -7)$

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(C) The equation of the tangent to the parabola $y = x^2 + 6$ at $(1, 7)$ is given by $\frac{1}{2}(y + 7) = x(1) + 6$.Simplifying this,we get $y + 7 = 2x + 12$,which implies $y = 2x + 5$ $(i)$.This tangent touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ $(ii)$.Substituting $(i)$ into $(ii)$:$x^2 + (2x + 5)^2 + 16x + 12(2x + 5) + c = 0$$x^2 + 4x^2 + 20x + 25 + 16x + 24x + 60 + c = 0$$5x^2 + 60x + (85 + c) = 0$.Since the line is tangent,the discriminant must be zero:$D = (60)^2 - 4(5)(85 + c) = 0$$3600 - 20(85 + c) = 0$$180 - (85 + c) = 0 \implies c = 95$.The quadratic equation becomes $5x^2 + 60x + 180 = 0$,or $x^2 + 12x + 36 = 0$.$(x + 6)^2 = 0 \implies x = -6$.Substituting $x = -6$ into $(i)$,$y = 2(-6) + 5 = -7$.Thus,the point of contact is $(-6, -7)$.

The tangent to the parabola $y = x^2 + 6$ at the point $(1, 7)$ touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ at which point?

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