Question

In the given figure, AB and CD are common tangents to two circles of unequal radii. if radii of the two circles are equal ,prove that AB=CD.

Answer 1

Given : Two circles of equal radii, two common tangents, AB and CD on circles, $C_1$ and $C_2$.

To prove : $AB=CD$

Construction : Join $O_{1}A,O_{1}C$ and $O_{2}B$ and $O_{2}D$ . Also join $O_1{O}_2$.

Proof : Since tangent at any point of a circle is perpendicular to the radius to the point of contact.

$\therefore \angle O_{1}AB=\angle O_{2}BA={90}^{\circ }$

As $O_{1}A=O_{2}B$, so $O_{1}AB{O}_2$ is a rectangle

Since opposite sides of a rectangle are equal

$\therefore AB=O_1{O}_2$ ___(i)

Similarly, we can prove that $O_{1}CD{O}_2$ is a rectangle.

$\therefore O_1{O}_2=CD$ ___(ii)

From (i) and (ii) , we get

$AB=CD$

Hence proved.