Question:
If \( S_1: x^2 + y^2 - 6x - 8y + 21 = 0 \) and \( S_2: x^2 + y^2 + 6x + 8y + \lambda = 0 \), then the distance of the centre of \( S_2 \) to the farthest point on \( S_1 \) is:
If \( S_1: x^2 + y^2 - 6x - 8y + 21 = 0 \) and \( S_2: x^2 + y^2 + 6x + 8y + \lambda = 0 \), then the distance of the centre of \( S_2 \) to the farthest point on \( S_1 \) is:
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For any point \( P \) and circle with centre \( C \) and radius \( r \), the nearest distance is \( |PC - r| \) and the farthest distance is \( PC + r \).
Updated On: Apr 6, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The farthest point on a circle from an external point lies on the line passing through the centre of the circle and that external point. The distance is calculated as the distance between the two centres plus the radius of the target circle.
Step 2: Key Formula or Approach:
1. Centre of \( S_1 \), \( C_1 = (-g_1, -f_1) \) and Radius \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} \). 2. Centre of \( S_2 \), \( C_2 = (-g_2, -f_2) \). 3. Farthest distance \( D_{max} = \text{dist}(C_1, C_2) + r_1 \).
Step 3: Detailed Explanation:
1. For \( S_1 \): \( g = -3, f = -4, c = 21 \). Centre \( C_1 = (3, 4) \). Radius \( r_1 = \sqrt{(-3)^2 + (-4)^2 - 21} = \sqrt{9 + 16 - 21} = \sqrt{4} = 2 \). 2. For \( S_2 \): \( g = 3, f = 4 \). Centre \( C_2 = (-3, -4) \). 3. Distance between centres \( C_1 \) and \( C_2 \): \( d = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10 \). 4. Farthest distance = \( d + r_1 = 10 + 2 = 12 \).
Step 4: Final Answer:
The distance to the farthest point is 12.
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