Question : P and Q are centre of two circles with radii 9 cm and 2 cm respectively, where PQ = 17 cm. R is the centre of another circle of radius $x$ cm, which touches each of the above two circles externally. If $\angle PRQ=90^{\circ}$, then the value of $x$ is:
Option 1: 4 cm
Option 2: 6 cm
Option 3: 7 cm
Option 4: 8 cm
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Correct Answer: 6 cm
Solution :
Given: PQ = 17 cm
PR = $(x+9)$ cm
RQ = $(x+2)$ cm
In right $\triangle$ PRQ
By Pythagoras theorem
$PQ^2 = PR^2+RQ^2$
⇒ $17^2 = (x+9)^2+(x+2)^2$
⇒ $289 = x^2+81+18x+x^2+4+4x$
⇒ $2x^2+22x-204=0$
⇒ $x^2+11x-102=0$
⇒ $x(x+17)-6(x+17)=0$
⇒ $(x+17)(x-6)=0$
$x = 6$ since $x$ cannot be negative.
Hence, the correct answer is $6$ cm.
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