Circle And System Of Circles Question 161
Question: The equation of the circle which touches x-axis at (3, 0) and passes through (1, 4) is given by
[MP PET 1993]
Options:
A) $ x^{2}+y^{2}-6x-5y+9=0 $
B) $ x^{2}+y^{2}+6x+5y-9=0 $
C) $ x^{2}+y^{2}-6x+5y-9=0 $
D) $ x^{2}+y^{2}+6x-5y+9=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ k^{2}=4+{{(k-4)}^{2}}\Rightarrow k=\frac{5}{2} $
Hence required equation of circle is
$ {{(x-3)}^{2}}+{{( y-\frac{5}{2} )}^{2}}={{( \frac{5}{2} )}^{2}}\Rightarrow x^{2}+y^{2}-6x-5y+9=0 $
Trick: Only (a) passes through (1, 4).
BETA
