Question:
The length of the tangent drawn from the point \( \left(\frac{k}{4}, \frac{k}{3}\right) \) to the circle \( x^2 + y^2 + 8x - 6y - 24 = 0 \) is:
The length of the tangent drawn from the point \( \left(\frac{k}{4}, \frac{k}{3}\right) \) to the circle \( x^2 + y^2 + 8x - 6y - 24 = 0 \) is:
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The length of a tangent from an external point to a circle is given by:
\[
L = \sqrt{h^2 + k^2 - r^2}
\]
Updated On: May 15, 2025
- \( 7 \)
- \( 1 \)
- \( 12 \)
- \( 24 \)
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Using the tangent length formula from a point \( (h, k) \) to a circle:
\[
L = \sqrt{h^2 + k^2 - r^2}
\]
After substituting values and solving, we obtain:
\[
L = 1
\]
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