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Two circles of radii r1 and r2 are both touching the coordinate axes and intersecting each other orthogonally. The value of r1/r2 (where r1 > r2) equals
- a)2 + √3
- b)√3 + 1
- c)2 – √3
- d)2 + √5
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Two circles of radii r1 and r2 are both touching the coordinate axes a...
Let the centers of the circles be A and B, with coordinates (r1, r1) and (-r2, -r2) respectively.
Since the circles are orthogonal, the distance between their centers is equal to the sum of their radii. Using the distance formula, we have:
√[(r1 - (-r2))^2 + (r1 - (-r2))^2] = r1 + r2
√[(r1 + r2)^2 + (r1 + r2)^2] = r1 + r2
√[2(r1 + r2)^2] = r1 + r2
√2(r1 + r2) = r1 + r2
√2 = 1 + r1/r2
Squaring both sides of the equation, we get:
2 = (1 + r1/r2)^2
2 = 1 + 2(r1/r2) + (r1/r2)^2
Rearranging the terms, we have:
(r1/r2)^2 + 2(r1/r2) - 1 = 0
Let x = r1/r2. Substituting this into the equation, we have:
x^2 + 2x - 1 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac))/2a
In this case, a = 1, b = 2, and c = -1. Plugging in these values, we get:
x = (-2 ± √(2^2 - 4(1)(-1)))/2(1)
x = (-2 ± √(4 + 4))/2
x = (-2 ± √8)/2
x = (-2 ± 2√2)/2
x = -1 ± √2
Since r1 and r2 are both positive, we take the positive value of x:
x = -1 + √2
Therefore, r1/r2 = -1 + √2.
Since the circles are orthogonal, the distance between their centers is equal to the sum of their radii. Using the distance formula, we have:
√[(r1 - (-r2))^2 + (r1 - (-r2))^2] = r1 + r2
√[(r1 + r2)^2 + (r1 + r2)^2] = r1 + r2
√[2(r1 + r2)^2] = r1 + r2
√2(r1 + r2) = r1 + r2
√2 = 1 + r1/r2
Squaring both sides of the equation, we get:
2 = (1 + r1/r2)^2
2 = 1 + 2(r1/r2) + (r1/r2)^2
Rearranging the terms, we have:
(r1/r2)^2 + 2(r1/r2) - 1 = 0
Let x = r1/r2. Substituting this into the equation, we have:
x^2 + 2x - 1 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac))/2a
In this case, a = 1, b = 2, and c = -1. Plugging in these values, we get:
x = (-2 ± √(2^2 - 4(1)(-1)))/2(1)
x = (-2 ± √(4 + 4))/2
x = (-2 ± √8)/2
x = (-2 ± 2√2)/2
x = -1 ± √2
Since r1 and r2 are both positive, we take the positive value of x:
x = -1 + √2
Therefore, r1/r2 = -1 + √2.
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Community Answer
Two circles of radii r1 and r2 are both touching the coordinate axes a...
Circle is (x – r)2 + (y – r)2 = r2
⇒ x2 + y2 – 2xr – 2yr + r2 = 0
Hence the circles are x2 + y2 – 2xr1 – 2yr1 + r12 = 0 ......(1)
x2 + y2 – 2xr2 – 2yr2 + r22 = 0 .....(2)
As (1) and (2) are orthogonal so
2r1r2 + 2r1r2 = r12 + r22
⇒ x2 + y2 – 2xr – 2yr + r2 = 0
Hence the circles are x2 + y2 – 2xr1 – 2yr1 + r12 = 0 ......(1)
x2 + y2 – 2xr2 – 2yr2 + r22 = 0 .....(2)
As (1) and (2) are orthogonal so
2r1r2 + 2r1r2 = r12 + r22
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Two circles of radii r1 and r2 are both touching the coordinate axes and intersecting each other orthogonally. The value of r1/r2(where r1 > r2) equalsa)2 + √3b)√3 + 1c)2 – √3d)2 + √5Correct answer is option 'A'. Can you explain this answer?
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