Question

Consider the circle \( x^2 + y^2 + 2gx + 2fy + 25 = 0 \) having centre on line \( -2x + y + 4 = 0 \) and radius 6. If the line \( x = 1 \) cuts the circle at A and B then \( AB^2 \) is:

• A. 40
• B. 60
• C. 80
• D. 100

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The center $(h, k)$ of the circle lies on $-2x + y + 4 = 0$, so $k = 2h - 4$. The radius is 6, which relates to the circle coefficients as $h^2 + k^2 - 25 = 6^2$.
Step 2: Key Formula or Approach
1. $h^2 + k^2 = 36 + 25 = 61$.
2. Distance $d$ from center $(h, k)$ to line $x=1$ is $|h-1|$.
3. Chord length $AB = 2\sqrt{r^2 - d^2}$.
Step 3: Detailed Explanation
Substitute $k = 2h - 4$ into $h^2 + k^2 = 61$:
$h^2 + (2h - 4)^2 = 61 \implies h^2 + 4h^2 - 16h + 16 = 61$
$5h^2 - 16h - 45 = 0 \implies (5h + 9)(h - 5) = 0$.
Since $h=5$ is a standard solution, $k = 2(5) - 4 = 6$. Center is $(5, 6)$.
Distance $d$ from $(5, 6)$ to $x=1$ is $|5-1| = 4$.
$AB = 2\sqrt{6^2 - 4^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$.
$AB^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$.
Step 4: Final Answer
The value of \( AB^2 \) is 80.