Question

Let \((\alpha, \beta)\) be the centre and \(\gamma\) be the radius of the circle \(x^2+y^2-6x-2y-15=0\), then the value of \((\alpha^2+\beta^2+\gamma^2)\) is:

• A. 9
• B. 35
• C. 21
• D. 42

Hint:

To find the center and radius from the general form \(x^2+y^2+2gx+2fy+c=0\), the center is \((-g, -f)\) and the radius is \(\sqrt{g^2+f^2-c}\). In this problem, \(2g=-6 \Rightarrow g=-3\), \(2f=-2 \Rightarrow f=-1\), \(c=-15\). Center is \((3,1)\), radius is \(\sqrt{9+1-(-15)} = \sqrt{25}=5\).

Answer 1

The Correct Option is B

Step 1: Convert the equation of the circle to standard form \((x-h)^2 + (y-k)^2 = r^2\).
We do this by completing the square for the x and y terms. \[ (x^2 - 6x) + (y^2 - 2y) = 15 \] To complete the square for x, add \((\frac{-6}{2})^2 = 9\). To complete the square for y, add \((\frac{-2}{2})^2 = 1\). \[ (x^2 - 6x + 9) + (y^2 - 2y + 1) = 15 + 9 + 1 \] \[ (x-3)^2 + (y-1)^2 = 25 \]

Step 2: Identify the center \((\alpha, \beta)\) and the radius \(\gamma\).
Comparing with the standard form, the center is \((h, k) = (\alpha, \beta) = (3, 1)\). The radius squared is \(r^2 = \gamma^2 = 25\), so the radius is \(\gamma = 5\).

Step 3: Calculate the required value \(\alpha^2+\beta^2+\gamma^2\). \[ \alpha^2+\beta^2+\gamma^2 = (3)^2 + (1)^2 + (5)^2 \] \[ = 9 + 1 + 25 = 35 \]