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:
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:
Updated On: Apr 8, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The center of the circle is $(-g, -f)$. Let the center be $(h, k)$. We know $(h, k)$ lies on $-2x + y + 4 = 0$, so $-2h + k + 4 = 0 \implies k = 2h - 4$. The radius is 6, so $h^2 + k^2 - 25 = 6^2 = 36$.
Step 2: Key Formula or Approach:
1. Radius formula: $g^2 + f^2 - c = r^2$. 2. Length of chord $AB = 2\sqrt{r^2 - d^2}$, where $d$ is the perpendicular distance from center to $x=1$.
Step 3: Detailed Explanation:
1. Substitute $k = 2h - 4$ into the radius equation: \[ h^2 + (2h-4)^2 - 25 = 36 \implies h^2 + 4h^2 - 16h + 16 - 25 = 36 \] \[ 5h^2 - 16h - 45 = 0 \implies (5h + 9)(h - 5) = 0 \] 2. For integer/standard solutions, take $h = 5$. Then $k = 2(5) - 4 = 6$. Center $C(5, 6)$. 3. Distance $d$ from $(5, 6)$ to line $x=1$: \[ d = |5 - 1| = 4 \] 4. Length $AB = 2\sqrt{r^2 - d^2} = 2\sqrt{6^2 - 4^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$. 5. $AB^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$.
Step 4: Final Answer:
The value of \( AB^2 \) is 80.
The center of the circle is $(-g, -f)$. Let the center be $(h, k)$. We know $(h, k)$ lies on $-2x + y + 4 = 0$, so $-2h + k + 4 = 0 \implies k = 2h - 4$. The radius is 6, so $h^2 + k^2 - 25 = 6^2 = 36$.
Step 2: Key Formula or Approach:
1. Radius formula: $g^2 + f^2 - c = r^2$. 2. Length of chord $AB = 2\sqrt{r^2 - d^2}$, where $d$ is the perpendicular distance from center to $x=1$.
Step 3: Detailed Explanation:
1. Substitute $k = 2h - 4$ into the radius equation: \[ h^2 + (2h-4)^2 - 25 = 36 \implies h^2 + 4h^2 - 16h + 16 - 25 = 36 \] \[ 5h^2 - 16h - 45 = 0 \implies (5h + 9)(h - 5) = 0 \] 2. For integer/standard solutions, take $h = 5$. Then $k = 2(5) - 4 = 6$. Center $C(5, 6)$. 3. Distance $d$ from $(5, 6)$ to line $x=1$: \[ d = |5 - 1| = 4 \] 4. Length $AB = 2\sqrt{r^2 - d^2} = 2\sqrt{6^2 - 4^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$. 5. $AB^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$.
Step 4: Final Answer:
The value of \( AB^2 \) is 80.
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