Question:
The equation of the circle which cuts all the three circles
\[
4(x-1)^2 + 4(y-1)^2 = 1,
4(x+1)^2 + 4(y-1)^2 = 1,
4(x+1)^2 + 4(y+1)^2 = 1,
\]
orthogonally is?
The equation of the circle which cuts all the three circles
\[
4(x-1)^2 + 4(y-1)^2 = 1,
4(x+1)^2 + 4(y-1)^2 = 1,
4(x+1)^2 + 4(y+1)^2 = 1, \] orthogonally is?
4(x+1)^2 + 4(y-1)^2 = 1,
4(x+1)^2 + 4(y+1)^2 = 1, \] orthogonally is?
Show Hint
Use the condition for two circles cutting orthogonally and solve simultaneous conditions.
Updated On: Jun 6, 2025
- \(4x^2 + 4y^2 = 49\)
- \(4(x-1)^2 + 4(y+1)^2 = 1\)
- \((x-1)^2 + (y+1)^2 = 4\)
- \(4x^2 + 4y^2 = 7\)
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The Correct Option is D
Solution and Explanation
Circles are given, and the required circle cuts all orthogonally.
Using orthogonality condition between circles:
\[
2 g g_1 + 2 f f_1 = c + c_1,
\]
solve to find required circle equation as
\[
4x^2 + 4y^2 = 7.
\]
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