Circle And System Of Circles Question 339
Question: If the middle point of a chord of the circle $ x^{2}+y^{2}+x-y-1=0 $ be (1, 1), then the length of the chord is
Options:
A) 4
B) 2
C) 5
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Here radius of circle is $ AC=\sqrt{{{( \frac{1}{2} )}^{2}}+{{( \frac{1}{2} )}^{2}}+1}=\sqrt{\frac{3}{2}} $ and $ CB=\sqrt{{{( \frac{3}{2} )}^{2}}+{{( \frac{1}{2} )}^{2}}}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}} $ Now $ AB^{2}=AC^{2}-BC^{2}=\frac{3}{2}-\frac{5}{2}=-1 $
$ \Rightarrow $ There is no possible value for AB. Aliter : Since the point (1, 1) lies outside the circle, therefore no such chord exist.
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