the tangent to curv y= x^2 + 6 at a point 1,7 touches the circle x^2+y^2+16x+12y+c=0 at a point q then the coordinates of q are :ans = -6,-7

The tangent to the curve $y={x}^{2}+6$ at point (1,7) is given by:

if equation (1) is the tangent to the circle ${x}^{2}+{y}^{2}+16x+12y+c=0$
substituting the value of y from eq(2),

since the line (1) touches the given circle, then discriminant of eq(3) must be zero:
${60}^{2}-4*5*\left(85+c\right)=0\phantom{\rule{0ex}{0ex}}3600-20*\left(85+c\right)=0\phantom{\rule{0ex}{0ex}}180-85-c=0\phantom{\rule{0ex}{0ex}}c=180-85=95\phantom{\rule{0ex}{0ex}}$
thus equation (3) can be rewritten as $5{x}^{2}+60x+85+95=0$
$⇒5{x}^{2}+60x+180=0\phantom{\rule{0ex}{0ex}}{x}^{2}+12x+36=0\phantom{\rule{0ex}{0ex}}⇒\left(x+6{\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x=-6$
for
$y=-12+5\phantom{\rule{0ex}{0ex}}y=-7$
if the point of contact is Q, then the coordinates of Q are $\left(-6,-7\right)$

hope this helps you

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