Question

In the figure P and Q are the centres of the circles with radii 9 cm and 2 cm respectively. If $\angle PRQ={90}^o$ and PQ = 17 cm find the radius of the circle with centre R.

Answer 1

Let the radius of the circle with centre $R=x$ units.

$\therefore$ In $△PQR,PQ=17cm$

$PR=(x+9)cm$

$QR=(x+2)cm$ and $\angle PRQ={90}^o$

$\therefore P{Q}^2=P{R}^2+Q{R}^2$ (Pythagoras theorem)

${17}^2=(x+9{)}^2+(x+2{)}^2$

$289=x^2+81+18x+x^2+4+4x$

$2x^2+22x+85-289=0$

$2x^2+22x-204=0$ [divided by 2]

$x^2+11x-102=0\Rightarrow (x+17)(x-6)=0$

$\therefore x+17=0$ or $x-6=0$

$x=-17$ or $x=6$

$\therefore$ radius of the circle with centre $R=6$ cm