Question:
The length of the tangent from the point \((5,1)\) to the circle
\[
x^2+y^2+6x-4y-3=0
\]
is
The length of the tangent from the point \((5,1)\) to the circle
\[
x^2+y^2+6x-4y-3=0
\]
is
Show Hint
For a circle \(S=0\), tangent length from \((x_1,y_1)\) is \(\sqrt{S_1}\).
Updated On: May 7, 2026
- \(81\)
- \(7\)
- \(29\)
- \(21\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: For circle \[ S=x^2+y^2+2gx+2fy+c=0 \] length of tangent from \((x_1,y_1)\) is: \[ \sqrt{S_1} \]
Step 2: Substitute \((5,1)\): \[ S_1=5^2+1^2+6(5)-4(1)-3 \] \[ =25+1+30-4-3 \] \[ =49 \]
Step 3: Length of tangent: \[ \sqrt{49}=7 \] \[ \boxed{7} \]
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