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) satisfyYou need to login to perform this action.
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