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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.Image
In $$\triangle ARB$$, 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 negative

1200230_1249850_ans_4fe8f52b9c8341b596f5e5dcd11bbf0f.jpg

Mathematics

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