A circle with radius $2$ is placed against a right angle. Another small circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

- $3-2 \sqrt{2}$
- $4-2 \sqrt{2}$
- $7-4 \sqrt{2}$
- $6-4 \sqrt{2}$

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A circle with radius $2$ is placed against a right angle. Another small circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

- $3-2 \sqrt{2}$
- $4-2 \sqrt{2}$
- $7-4 \sqrt{2}$
- $6-4 \sqrt{2}$

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Best answer

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If we draw a line joining the centres of the two circles and leading to the intersection point of the two tangents , that can help us arrive at the solution.

Let r be the radius of the smaller circle so distance between centre and the point of intersection of tangent = sqrt(r^{2} + r^{2}) = √2 r.

And similarly distance between centre of larger circle and the point of intersection of tangent = √ (2^{2 }+ 2^{2}) = 2√2 r

Also this distance can be written as : radius of larger circle + radius of smaller circle + distance between centre of smaller circle to point of intersection of tangents = 2 + r + √2 r.

Hence we have :

2 + r + √2 r = 2√2

==> r(1 + √2) = 2(√2 - 1)

==> r = 2(√2 - 1) / (√2 + 1)

==> r = [ 2(√2 - 1) / (√2 + 1) ] * [ (√2 - 1) / (√2 - 1) ]

= 2(√2 - 1)^{2} / [ (√2)^{2} - (1)^{2} ]

= 2(√2 - 1)^{2}

= 6 - 4√2

**Hence 4) should be the correct answer.**

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