The ratio of length of direct common tangents to the length of transverse common tangents for the circle $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 4 x - 6 y + 12 = 0$ is $\sqrt{k} / 3$ then k =

Open in App

Solution

$\begin{matrix} x^{2} + y^{2} = 1 & x^{2} + y^{2} - 4 x - b y + 12 = 0 C_{1} (0, 0) & C_{2} (2, 3) r = 1 & r = \sqrt{4 + 9 - 12} = 1 c_{1} c_{2} = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13} r_{1} + r_{2} = 2 c_{1} c_{2} > r_{1} + r_{2} \end{matrix}$
Then 2 direct tangent
2 transverse tangent
$\Rightarrow$ Length of direct common tangent $= \sqrt{{(c_{1} c_{2})}^{2} + {(r_{1} + r_{2})}^{2}} = \sqrt{13 + 0} = \sqrt{13}$
$\Rightarrow$ Length of transverse common tangent $= \sqrt{{(c_{1} c_{2})}^{2} - {(r_{1} + r_{2})}^{2}} = \sqrt{13 - 2^{2}} = \sqrt{9} = 3$
$\Rightarrow$ Then ratio of direct to transverse tangent $= \frac{\sqrt{13}}{3}$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

x^{2} + y^{2} + 6 x + 4 y + 4 = 0

x^{2} + y^{2} - 2 x = 0

y - 1 = 0

4 x - 3 y - 9 = 0

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

x - 1 = 0, 3 x + 4 y - 21 = 0

Find the Slope of the transverse common tangent to the circles $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

Join BYJU'S Learning Program