Question

The power of a point (2,-1) with respect to a circle C of radius 4 is 9. The centre of the circle C lies on the line x+y=0 and in the 2nd quadrant. If \((\alpha, \beta)\) is the centre of the circle C, then \(\beta - \alpha = \)

• A. -4
• B. -10
• C. 4
• D. 10

Hint:

The power of a point \(P(x_1, y_1)\) with respect to the circle \(x^2+y^2+2gx+2fy+c=0\) is simply \(S_1 = x_1^2+y_1^2+2gx_1+2fy_1+c\). Geometrically, it's also equal to \(d^2 - r^2\), where d is the distance from the point to the center.

Answer 1

The Correct Option is C

Let the center of the circle C be \((\alpha, \beta)\).
The power of a point P with respect to a circle is defined as \(d^2 - r^2\), where d is the distance from P to the center and r is the radius.
We are given the point P(2,-1), power = 9, and radius r = 4.
The distance \(d\) is the distance between P(2,-1) and the center \((\alpha, \beta)\).
\(d^2 = (2-\alpha)^2 + (-1-\beta)^2\).
Using the power formula: \( (2-\alpha)^2 + (-1-\beta)^2 - 4^2 = 9 \).
\( (2-\alpha)^2 + (1+\beta)^2 - 16 = 9 \implies (2-\alpha)^2 + (1+\beta)^2 = 25 \). (Equation 1)
The center \((\alpha, \beta)\) lies on the line \(x+y=0\), so \( \alpha + \beta = 0 \implies \beta = -\alpha \). (Equation 2)
The center is in the 2nd quadrant, which means \( \alpha<0 \) and \( \beta>0 \).
Substitute \(\beta = -\alpha\) into Equation 1.
\( (2-\alpha)^2 + (1-\alpha)^2 = 25 \).
\( (4 - 4\alpha + \alpha^2) + (1 - 2\alpha + \alpha^2) = 25 \).
\( 2\alpha^2 - 6\alpha + 5 = 25 \).
\( 2\alpha^2 - 6\alpha - 20 = 0 \implies \alpha^2 - 3\alpha - 10 = 0 \).
Factoring the quadratic: \( (\alpha-5)(\alpha+2) = 0 \).
The possible values for \(\alpha\) are \(\alpha=5\) or \(\alpha=-2\).
Since the center is in the 2nd quadrant, we must have \( \alpha<0 \). So we choose \( \alpha = -2 \).
Now find \(\beta\) using \(\beta = -\alpha\).
\( \beta = -(-2) = 2 \).
The center is \((-2, 2)\), which is indeed in the 2nd quadrant.
Finally, we need to calculate \( \beta - \alpha \).
\( \beta - \alpha = 2 - (-2) = 4 \).