Conic-Sections Question 584
Question: The equation of the circle which touches the axes at a distance 5 from the origin is $ y^{2}+x^{2}-2ax-2ay+a^{2}=0. $ what is the value of $ \alpha $ ?
Options:
A) 4
B) 5
C) 6
D) 7
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Coordinates of the centre of given circle $ =(\alpha ,\alpha ) $ and radius $ =\sqrt{{{(\alpha )}^{2}}+{{(\alpha )}^{2}}-{{\alpha }^{2}}}=\sqrt{{{\alpha }^{2}}}=\alpha $
$ \therefore {{(\alpha -5)}^{2}}+{{(\alpha )}^{2}}={{(\alpha )}^{2}} $
$ \Rightarrow {{\alpha }^{2}}+25-10\alpha =0\Rightarrow {{(\alpha -5)}^{2}}=0\Rightarrow \alpha =5 $ Then, other root will always real.
BETA
