Question

Two circles of radii $x$ and $y$ touch each other externally and $ST$ is their common tangent, touching them at $S$ and $T$. Then $ST$ ${}^2$ equals -

• A. $x-y$
• B. $x+y$
• C. $2xy$
• D. $4xy$

Answer 1

The correct option is D $4xy$

$\because$ Tangent is perpendicular to normal (line joining point of contact of tangent and centre of circle).

$Hence,LOST=L{O}^{\prime }TS={90}^0.\therefore OS\parallel OTExtendT{O}^{1}suchthatTR\parallel SOIfTR\parallel SO,\therefore TSORisarectangle.\therefore TR=SO\Rightarrow andST=OR\to \left(1\right).Now,in△O{O}^{1}RO{R}^2+R{O}^{12}=O{O}^{1}From\left(D\right)\to S{T}^2+R{O}^1=O{O}^{12}\Rightarrow S{T}^2=O{O}^{12}-R{O}^{12}R{O}^1=RT-TO=x-y.andO{O}^1=x+y.\Rightarrow S{T}^2=(x+y{)}^2-(x-y{)}^2=x^2+y^2+2xy-(x^2+y^2-2xy)=4xy.\therefore D)Ans.$