2017-I-13 – Solving Master

Ans: (a)

(x - 2)^{2} + (y + 1)^{2} = 100

F

G

and

H

are collinear. (ii)

12 x + 5 y - 19 = 0

Note that the radius of $C$
$\begin{array}{cl} = & \sqrt{(- 6 - 2)^{2} + (5 - (- 1))^{2}} \\ = & 10 \end{array}$

Hence, the equation of $C$ is

$\begin{array}{rcl} (x - 2)^{2} + (y - (- 1))^{2} & = & 10^{2} \\ x^{2} - 4 x + 4 + y^{2} + 2 y + 1 - 100 & = & 0 \\ x^{2} + y^{2} - 4 x + 2 y - 95 & = & 0 \end{array}$
The distance between $F$ and $G$
$\begin{array}{cl} = & \sqrt{(- 3 - 2)^{2} + (11 - (- 1))^{2}} \\ = & 13 \\ > & 10, which is the radius of C \end{array}$

Therefore, $F$ lies outside $C$ .
1. $F$ , $G$ and $H$ are collinear.
2. The required equation is
  $\begin{array}{rcl} \frac{y - (- 1)}{x - 2} & = & \frac{11 - (- 1)}{- 3 - 2} \\ - 5 y - 5 & = & 12 x - 24 \\ 12 x + 5 y - 19 & = & 0 \end{array}$