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If length of the tangent from origin to the circle \( \Large x^{2}+y^{2}-26x+K=0\ is\ 5 \), then K is equal to

A) \( \Large \sqrt{5} \)
B) 5
C) 10
D) 25

Correct Answer:




D) 25

Description for Correct answer:

Given circle is

\( \Large \left(x-13\right)^{2}+y^{2}=169-K \)

Therefore, Centre is \( \Large \left(13,\ 0\right)\ and\ radius = \sqrt{169-K} \)

From \( \Large \triangle OCT,\ OC^{2}=OT^{2}+CT^{2} \)

Therefore, \( \Large 13^{2}=5^{2}+169-K \)

or K = 25

Part of solved Loci and concurrency questions and answers : >> Elementary Mathematics >> Loci and concurrency

Comments

Similar Questions

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