If the tangent at (1, 7) to the curve x2 = y - 6
touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is :
Solutions · 1
Official solution
NOTE :
Equation of tangent at (x1, y1) to the curve x2 = 4ay is
xx1 = 4a \[\left( {{{y + {y_1}} \over 2}} \right)\]
Now equation of tangent at (1, 7) to x2 = y - 6 is
\[x\,.\,1 = 4\,.\,{1 \over 4}\left( {{{y + 7} \over 2}} \right) - 6\]
\[ \Rightarrow 2x = y + 7 - 12\]
\[ \Rightarrow 2x - y + 5 = 0\]
This tangent touches the circle.
So, perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.
For the circle,
x2+y2+16x+12y+C=0
center is (-8, -6)
and radius (r) = \[\sqrt {{8^2} + {6^2} - c} \]
\[ = \sqrt {100 - c} \]
Distance of the tangent from the center of the circle
d = \[\left| {{{2\left( { - 8} \right) - \left( { - 6} \right) + 5} \over {\sqrt {{2^2} + {1^2}} }}} \right|\]
\[ = \left| {{{ - 16 + 6 + 5} \over {\sqrt 5 }}} \right|\]
And we know d = r
\[\therefore\,\,\,\] \[\left| {{{ - 16 + 11} \over {\sqrt 5 }}} \right| = \sqrt {100 - c} \]
\[ \Rightarrow \left| {{{ - 5} \over {\sqrt 5 }}} \right| = \sqrt {100 - c} \]
\[ \Rightarrow \left| { - \sqrt 5 } \right| = \sqrt {100 - c} \]
\[ \Rightarrow 5 = 100 - c\]
\[ \Rightarrow c = 95\]