Question

If the length of an internal tangent of two circles is 7 and the length of an external tangent is 11, then the product of radii of two circles is
[a] 18
[b] 20
[c] 16
[d] 12

Answer 1

Hint:Use the property that the length of an internal tangent $=\sqrt{{{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}}$ and the length of an external tangent $=\sqrt{{{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}}$, where ${{C}_{1}}{{C}_{2}}$ is the distance between the centres of the two circles,${{r}_{1}}$ is the radius of one circle and ${{r}_{2}}$is the radius of the other circle. Form two equations using the above results and the statement of the question and eliminate ${{C}_{1}}{{C}_{2}}$to get the result.

Complete step-by-step answer:
Let ${{r}_{1}}$ be the radius of one circle and ${{r}_{2}}$ the radius of another circle. Let d be the distance between the centres of the circle.
Now we know that the length of an internal tangent $=\sqrt{{{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}}$
Since the length of an internal tangent = 7, we have
$\sqrt{{{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}}=7$
Squaring both sides, we get
$\begin{aligned} & {{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}={{7}^{2}} \\\ & \Rightarrow {{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}=49\text{ (i)} \\\ \end{aligned}$
Also, we know that length of an external tangent $=\sqrt{{{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}}$
Since the length of an external tangent = 11, we have
$\sqrt{{{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}}=11$
Squaring both sides, we get
${{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}=121\text{ (ii)}$
Subtracting equation (i) from equation (ii), we get
$\begin{aligned} & {{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}-\left( {{C}_{1}}{{C}_{2}}^{2}-{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}} \right)=121-49 \\\ & \Rightarrow {{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}-{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}=72 \\\ \end{aligned}$
Using ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get

& \left( {{r}_{1}}+{{r}_{2}}+{{r}_{1}}-{{r}_{2}} \right)\left( {{r}_{1}}+{{r}_{2}}-\left( {{r}_{1}}-{{r}_{2}} \right) \right)=72 \\\ & \Rightarrow 2{{r}_{1}}\left( 2{{r}_{2}} \right)=72 \\\ & \Rightarrow {{r}_{1}}{{r}_{2}}=18 \\\ \end{aligned}$$ Hence the product of radii = 18 Hence option [b] is correct Note: [1] Alternatively use Product of radii $=\dfrac{{{L}^{2}}-{{l}^{2}}}{4}$, where L is the length of direct common tangent and l is the length of indirect common tangent. Here L = 11 and l = 7 . Hence Product of radii $=\dfrac{{{L}^{2}}-{{l}^{2}}}{4}=\dfrac{121-49}{4}=\dfrac{72}{4}=18$