Search any question & find its solution
Question:
Answered & Verified by Expert
Let 6,8 be the $\mathrm{X}$ and $\mathrm{Y}$ - intercepts made by the circle $\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ respectively. If $\mathrm{gx}+\mathrm{fy}+1$ $=0$ is a line passing through the point $(1,-1)$, then the radius of the circle $S=0$ is
Options:
Solution:
2133 Upvotes
Verified Answer
The correct answer is:
$5$
$2 \sqrt{g^2-c}=6 \Rightarrow g^2-c=9$ ...(1)
$2 \sqrt{f^2-c}=8 \Rightarrow f^2-c=16$ ...(2)
$g x+f y+1=0$ passes through $(1,-1)$
$\therefore g-f+1=0 \Rightarrow g-f=-1$ ...(3)
Equations (1) - (2), $g^2-f^2=-7$
$\Rightarrow(g+f)(g-f)=-7 \Rightarrow g+f=7$ ...(4)
From (3) and (4), we get $g=3, f=4$
$$
\begin{array}{ll}
\therefore & c=g^2-9=0 \\
\therefore & \text { Radius }=\sqrt{g^2+f^2-c}=\sqrt{3^2+4^2}=5 .
\end{array}
$$
$2 \sqrt{f^2-c}=8 \Rightarrow f^2-c=16$ ...(2)
$g x+f y+1=0$ passes through $(1,-1)$
$\therefore g-f+1=0 \Rightarrow g-f=-1$ ...(3)
Equations (1) - (2), $g^2-f^2=-7$
$\Rightarrow(g+f)(g-f)=-7 \Rightarrow g+f=7$ ...(4)
From (3) and (4), we get $g=3, f=4$
$$
\begin{array}{ll}
\therefore & c=g^2-9=0 \\
\therefore & \text { Radius }=\sqrt{g^2+f^2-c}=\sqrt{3^2+4^2}=5 .
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.