Question:
The tangent to the circle \(x^2+y^2 = 10\) at the point \((3,1)\) touches the circle \(x^2+y^2-2\sqrt{10}\,x-20y+k = 0\), then the value of \(k\) is...
The tangent to the circle \(x^2+y^2 = 10\) at the point \((3,1)\) touches the circle \(x^2+y^2-2\sqrt{10}\,x-20y+k = 0\), then the value of \(k\) is...
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The tangent is a tangent to the second circle too, so its distance from that centre equals its radius.
Updated On: Oct 1, 2026
- \(-109\)
- \(109\)
- \(-101\)
- \(101\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
A line touches a circle when the distance from the centre to the line equals the radius.
Step 2: Key Formula or Approach:
Tangent to \(x^2 + y^2 = 10\) at \((3,1)\): \(3x + y = 10\). Second circle: centre \((\sqrt{10}, 10)\), radius\(^2 = (\sqrt{10})^2 + 10^2 - k = 110 - k\).
Step 3: Detailed Explanation:
Distance of the centre from \(3x + y - 10 = 0\):
\[ d = \frac{|3\sqrt{10} + 10 - 10|}{\sqrt{9 + 1}} = \frac{3\sqrt{10}}{\sqrt{10}} = 3 \]
Set \(r = d\): \(\sqrt{110 - k} = 3\), so \(110 - k = 9\).
\[ k = 101 \]
Final Answer:
\(k = 101\), option (D). \[ \boxed{101} \]
A line touches a circle when the distance from the centre to the line equals the radius.
Step 2: Key Formula or Approach:
Tangent to \(x^2 + y^2 = 10\) at \((3,1)\): \(3x + y = 10\). Second circle: centre \((\sqrt{10}, 10)\), radius\(^2 = (\sqrt{10})^2 + 10^2 - k = 110 - k\).
Step 3: Detailed Explanation:
Distance of the centre from \(3x + y - 10 = 0\):
\[ d = \frac{|3\sqrt{10} + 10 - 10|}{\sqrt{9 + 1}} = \frac{3\sqrt{10}}{\sqrt{10}} = 3 \]
Set \(r = d\): \(\sqrt{110 - k} = 3\), so \(110 - k = 9\).
\[ k = 101 \]
Final Answer:
\(k = 101\), option (D). \[ \boxed{101} \]
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