Question

The radius of two circles are $3.2$cm and $1.5$cm. The distance between centres is $6.2$cm. Construct a common transversal tangents. Find the length of tangents.

Answer 1

In the figure above,

$O_1{O}_2=6.2cm$

$O_{1}R=1.5cm$

$O_{2}S=3.2cm$

$PQ$ and $RS$ are the transversal tangents and $PQ=RS$

Since the radii are perpendicular to tangents, $\angle O_{1}RT=\angle O_{2}ST={90}^{\circ }$

Consider the two triangles, $\Delta O_{1}RT$ and $\Delta O_{2}ST$:

$\angle O_{1}RT=\angle O_{2}ST={90}^{\circ }$

$\angle RT{O}_1=\angle ST{O}_2\left(VerticallyOppositeAngles\right)$

Thus, $\Delta O_{1}RT\sim \Delta O_{2}ST$ by AA similarity criterion

Hence, $\frac{O_{1}T}{O_{2}T}=\frac{RT}{ST}=\frac{1.5}{3.2}$.

Therefore, $O_{1}T+\frac{1.5}{3.2}\times O_{1}T=6.2$

$⟹O_{1}T=4.22$

$O_{2}T=1.5/3.2\times 4.22$

$⟹O_{2}T=1.98$

By Pythagoras theorem, $RT=\sqrt{O_1{R}^2+O_1{T}^2}=\sqrt{{1.5}^2+{4.22}^2}$

$⟹RT=4.48$

Similarly, $ST=\sqrt{O_2{S}^2+O_2{T}^2}=\sqrt{{3.2}^2+{1.98}^2}$

$⟹ST=3.76$

Thus, $RS=RT+ST=4.48+3.76$

$⟹RS=PQ=8.24cm$