Question

If the line x+y=2 cuts the circle \(x^2+y^2+2x-4y+4=0\) at two points A and B then the radius of the circle passing through A, B and orthogonal to \(x^2+y^2-2x-4y-4=0\) is

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

The equation of any curve passing through the intersection of two curves \(U=0\) and \(V=0\) can be written as \(U+\lambda V=0\). For circles, this is \(S+\lambda L=0\) for a circle and a line, and \(S_1+\lambda S_2=0\) for two circles.

Answer 1

The Correct Option is C

Let S be \(x^2+y^2+2x-4y+4=0\) and L be \(x+y-2=0\).
The equation of any circle passing through the intersection of S and L is given by the family of circles \(S + \lambda L = 0\).
So the required circle, let's call it \(S_{req}\), has the equation:
\( (x^2+y^2+2x-4y+4) + \lambda(x+y-2) = 0 \).
\( S_{req}: x^2+y^2 + (2+\lambda)x + (-4+\lambda)y + (4-2\lambda) = 0 \).
This circle is orthogonal to the circle \(S' : x^2+y^2-2x-4y-4=0\).
The condition for orthogonality of two circles \(x^2+y^2+2g_1x+2f_1y+c_1=0\) and \(x^2+y^2+2g_2x+2f_2y+c_2=0\) is \(2g_1g_2 + 2f_1f_2 = c_1+c_2\).
For \(S_{req}\): \(g_1 = \frac{2+\lambda}{2}\), \(f_1 = \frac{-4+\lambda}{2}\), \(c_1 = 4-2\lambda\).
For \(S'\): \(g_2 = -1\), \(f_2 = -2\), \(c_2 = -4\).
Applying the condition: \( 2\left(\frac{2+\lambda}{2}\right)(-1) + 2\left(\frac{-4+\lambda}{2}\right)(-2) = (4-2\lambda) + (-4) \).
\( -(2+\lambda) - 2(-4+\lambda) = -2\lambda \).
\( -2 - \lambda + 8 - 2\lambda = -2\lambda \).
\( 6 - 3\lambda = -2\lambda \implies \lambda = 6 \).
Now substitute \( \lambda=6 \) back into the equation for \(S_{req}\).
\( x^2+y^2 + (2+6)x + (-4+6)y + (4-2(6)) = 0 \).
\( x^2+y^2 + 8x + 2y - 8 = 0 \).
Finally, find the radius of this circle.
Radius \(r = \sqrt{g^2+f^2-c} = \sqrt{4^2 + 1^2 - (-8)} = \sqrt{16+1+8} = \sqrt{25} = 5 \).