The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and…
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
- \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
- \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
- \((0,-1)\)
- \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
Solution
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, \(P(x, y)=(0,-1)\)
Asked in: AP EAMCET 2019 (22 Apr Shift 1)
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