Search any question & find its solution
Question:
Answered & Verified by Expert
If the length of the tangent from $(f, g)$ to the circle $x^2+y^2=6$ be twice the length of the tangent from the same point to the circle $x^2+y^2+3 x+3 y=0$, then $f^2+g^2+4 f+4 g+2$ is equal to
Options:
Solution:
1901 Upvotes
Verified Answer
The correct answer is:
0
$$
\begin{aligned}
& \text { Given, } P=(f, g) \\
& S: x^2+y^2-6=0 \\
& S^{\prime}: x^2+y^2+3 x+3 y=0
\end{aligned}
$$
According to the question,
$$
\sqrt{S_{11}}=2 \sqrt{S_{11}^{\prime}}
$$
Squaring on both sides
$$
\left(\sqrt{S_{11}}\right)^2=\left(2 \sqrt{S_{11}^{\prime}}\right)^2
$$
$$
\begin{gathered}
S_{11}=4 S_{11}^{\prime} \\
\left(g^2+f^2-6\right)=4\left(g^2+f^2+3 g+3 f\right) \\
3 g^2+3 f^2+12 g+12 f+6=0
\end{gathered}
$$
Dividing by 3 on both sides,
$$
g^2+f^2+4 g+4 f+2=0
$$
Hence, option (3) is correct.
\begin{aligned}
& \text { Given, } P=(f, g) \\
& S: x^2+y^2-6=0 \\
& S^{\prime}: x^2+y^2+3 x+3 y=0
\end{aligned}
$$
According to the question,
$$
\sqrt{S_{11}}=2 \sqrt{S_{11}^{\prime}}
$$
Squaring on both sides
$$
\left(\sqrt{S_{11}}\right)^2=\left(2 \sqrt{S_{11}^{\prime}}\right)^2
$$
$$
\begin{gathered}
S_{11}=4 S_{11}^{\prime} \\
\left(g^2+f^2-6\right)=4\left(g^2+f^2+3 g+3 f\right) \\
3 g^2+3 f^2+12 g+12 f+6=0
\end{gathered}
$$
Dividing by 3 on both sides,
$$
g^2+f^2+4 g+4 f+2=0
$$
Hence, option (3) is correct.
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.