Question

A line with gradient $2$ is passing through the point $P(1,7)$ and touches the circle $x^2+y^2+16x+12y+c=0$ at the point $Q$. If $(a,b)$ are the coordinates of the point $Q$, then find the value of $(7a+7b+c)$.

Answer 1

$\to$ given equation

$x^2+y^2+16x+12y+c=0$

standard circle equation

$\therefore x^2+y^2+2gx+2fy+c=0$

$2g=16$ $2f=12$

$g=8$ $f=6$

centre of circle $=(-g,-f)$

$=(-8,-6)$

$\to$ line passing from $P$ & $Q$

$\therefore m=2$

$\therefore y-y_1=m(x-x_1)$

$\therefore y-7=2(x-1)$

$\therefore y-7=2x-2$

$\therefore y=ex+5$

$\therefore b=29+5---\left(1\right)$

radius of circle

$=\sqrt{g^2+f^2-c}$

$=\sqrt{(8{)}^2+(6{)}^2-C}$

$=\sqrt{64+36-C}$

$r=\sqrt{100-C}$

$\to$ line passing from centre & $Q$

$\therefore m_1.m_2=-1$

$\therefore m_1.2=-1$

$m_1=\frac{-1}{2}$

$\therefore \frac{b-(-6)}{a-(-8)}=\frac{-1}{2}$

$\therefore \frac{b+6}{a+8}=\frac{-1}{2}$

$\therefore 2b+12=-a-8$

$\therefore 4a+10+12=-a-8$

$\therefore 5a=-8-12-10$

$\therefore 54=-30$

$\therefore a=-6$

Put value of $a$ in equation $\left(ii\right)$

$\therefore b=2(-6)+5$

$=-12+5$

$\therefore b=-7$

distance between $C$ & $Q$

$\to CQ=\sqrt{x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(-6-(-8{)}^2+(-7-(-6{)}^2}=\sqrt{(6-+8{)}^2+(-7+6{)}^2}=\sqrt{2^2+(-1{)}^2}=\sqrt{4+1}=\sqrt{5}$

$\to$ Radius $R=CQ=\sqrt{100-C}=\sqrt{5}$ $\therefore 100-C=5C=95cm$

$\to (7a+7b+7c)=7(-6-7+95)$

$=574$