Circle And System Of Circles Question 353
Question: Line $ y=x+a\sqrt{2} $ is a tangent to the circle $ x^{2}+y^{2}=a^{2} $ at
[RPET 1991; MP PET 1999]
Options:
A) $ ( \frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}} ) $
B) $ ( -\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}} ) $
C) $ ( \frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}} ) $
D) $ ( -\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
Suppose that the point be (h, k). Tangent at (h, k) is $ hx+ky=a^{2}\equiv x-y=-\sqrt{2}a $ or $ \frac{h}{1}=\frac{k}{-1}=\frac{a^{2}}{-\sqrt{2}a} $ or $ h=-\frac{a}{\sqrt{2}},\ k=\frac{a}{\sqrt{2}} $ Therefore, point of contact is $ ( -\frac{a}{\sqrt{2}},\ \frac{a}{\sqrt{2}} ) $ .
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