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Q. The total number of common tangents of $x^2 + y^2 - 6x -8y + 9 = 0$ and $x^2 + y^2 = 1$ is

KCETKCET 2011Conic Sections

Solution:

The given equation of circles
$x^{2}+y^{2}-6 x-8 y+9=0$
and $x^{2}+y^{2}=1$
Radius and centre of both circles is
$C_{1} \rightarrow(3,4), R_{1}=\sqrt{9+16-9}=4$
$C_{2} \rightarrow (0,0), R_{2}=1$
$C_{1} C_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$
$R_{1}+R_{2}=4+1=5$
$\because C_{1} C_{2}=R_{1}+R_{2}$
In this case, two direct tangent are real and distinct while the transverse tangents are coincident.
So, number of comman tangents $=3$