question_answer

The circle passing through (1, −2) and touching the axis of\[x\]at (3, 0) also passes through the point :     AIEEE Solevd Paper-2013

A) (−5, 2)                  

B) (2, −5)  

C)        (5, −2)  

D)        (−2, 5)

Correct Answer: C

Solution :

Let centre C(3, k) As touches X−axis \[\Rightarrow \] \[r=k\] So, circle is \[{{(x-3)}^{2}}+{{(y-k)}^{2}}={{k}^{2}}\] Given it passes (1, −7) \[4+{{(k+2)}^{2}}={{k}^{2}}\] \[4+{{k}^{2}}+4k+4={{k}^{2}}\] \[4k=-8\] \[k=-2\] Circle is \[{{(x-3)}^{2}}+{{(y+2)}^{2}}=4\] Obviously (5, −2) satisfy