Question

# $$A$$ & $$B$$ are centres of circles of radii $$9$$cm and $$2$$cm respectively. $$AB=17\ cm$$. $$R$$ is the center of the circle of radius $$x\ cm$$ which touches the above circle externally. Given that angle $$ARB$$ is $$90^{o}$$. Write an equation in $$x$$ and solve it.

Solution

## REF.ImageIn $$\triangle ARB$$, by Pythagoras theorem$$AB^{2}= AR^{2}+RB^{2}$$$$17^{2}=(x+9)^{2}+(x+2)^{2}$$$$289= x^{2}+81+18x+x^{2}+4+4x$$$$289 = 2x^{2}+22x+85$$$$2x^{2}+22x- 204=0$$$$x^{2}+11x-102=0$$$$x^{2}+17x-6x- 102=0$$$$x(x+17)+6(x+17)=0$$$$(x+17)(x-6)=0$$$$x=6,-17$$$$\therefore x=6$$ as cannot be negativeMathematics

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