Question

If a circle of radius \(3\) passes through the point \((7,3)\) and has its centre on the line \[ x-y-1=0, \] then its equation among the following is:

• A. \(x^2+y^2+14x-12y+76=0\)
• B. \(x^2+y^2+14x-12y+76=0\)
• C. \(x^2+y^2+8x-6y+16=0\)
• D. \(x^2+y^2-14x-12y+76=0\)

Hint:

For \(x^2+y^2+2gx+2fy+c=0\), the centre is \((-g,-f)\) and the radius is \(\sqrt{g^2+f^2-c}\).

Answer 1

The Correct Option is D

Step 1: Let the centre be (h, k) and apply the line condition.
Since the centre lies on x - y - 1 = 0, we have h - k = 1.

Step 2: Apply the radius condition using the given point.
The circle has radius 3 and passes through (7, 3). The distance equation gives (h - 7)² + (k - 3)² = 9.

Step 3: Identify the centre from the candidate equation.
For the circle x² + y² - 14x - 12y + 76 = 0, comparing with x² + y² + 2gx + 2fy + c = 0 gives 2g = -14 → g = -7 and 2f = -12 → f = -6. The centre is (-g, -f) = (7, 6).

Step 4: Verify both conditions.
Check the line: 7 - 6 - 1 = 0, satisfied. Check the radius: distance from (7, 6) to (7, 3) is √[(0)² + (3)²] = 3, which matches.

Step 5: Final conclusion.
The required circle is x² + y² - 14x - 12y + 76 = 0.