Search any question & find its solution
Question:
Answered & Verified by Expert
The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and \(x^2+y^2-6 x-4 y+4=0\), is
Options:
Solution:
2394 Upvotes
Verified Answer
The correct answer is:
\((0,-1)\)
Given equation of circles are
\(\begin{aligned}
& x^2+y^2-4 x-2 y+1=0 \quad \ldots (i) \\
& \text{and } x^2+y^2-6 x-4 y+4=0 \quad \ldots (ii)
\end{aligned}\)
Here,
\(x^2+y^2-6 x-4 y+4=0\)
\(\begin{aligned}
& C_1=(2,1), C_2=(3,2) \\
& r_1=\sqrt{4+1-1}=\sqrt{4}=2 \\
& \text{and } r_2=\sqrt{9+4-4}=\sqrt{9}=3
\end{aligned}\)
and \(r_2=\sqrt{9+4-4}=\sqrt{9}=3\)
Now, \(\begin{aligned}
C_1 C_2 & =\sqrt{(3-2)^2+(2-1)^2} \\
& =\sqrt{1+1}=\sqrt{2}
\end{aligned}\)
\(\begin{array}{ll}
\text {and } & r_1+r_2=2+3=5 \\
\therefore & C_1 C_2 < r_1+r_2
\end{array}\)
So, circles intersect at two distinct point.
let \(P(x, y)\) be the point of intersection of tangents
\(\therefore \quad P(x, y)=\left(\frac{6-6}{-1}, \frac{4-3}{-1}\right)\)
Hence, \(P(x, y)=(0,-1)\)
\(\begin{aligned}
& x^2+y^2-4 x-2 y+1=0 \quad \ldots (i) \\
& \text{and } x^2+y^2-6 x-4 y+4=0 \quad \ldots (ii)
\end{aligned}\)
Here,
\(x^2+y^2-6 x-4 y+4=0\)
\(\begin{aligned}
& C_1=(2,1), C_2=(3,2) \\
& r_1=\sqrt{4+1-1}=\sqrt{4}=2 \\
& \text{and } r_2=\sqrt{9+4-4}=\sqrt{9}=3
\end{aligned}\)
and \(r_2=\sqrt{9+4-4}=\sqrt{9}=3\)
Now, \(\begin{aligned}
C_1 C_2 & =\sqrt{(3-2)^2+(2-1)^2} \\
& =\sqrt{1+1}=\sqrt{2}
\end{aligned}\)
\(\begin{array}{ll}
\text {and } & r_1+r_2=2+3=5 \\
\therefore & C_1 C_2 < r_1+r_2
\end{array}\)
So, circles intersect at two distinct point.
let \(P(x, y)\) be the point of intersection of tangents
\(\therefore \quad P(x, y)=\left(\frac{6-6}{-1}, \frac{4-3}{-1}\right)\)
Hence, \(P(x, y)=(0,-1)\)
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.