Question 146 of 219MediumMCQ
The circle passing through (1,−2) and touching the axis of x at (3,0) also passes through the point: Solutions · 1
Official solution
Since circle touches x-axis at (3,0)
\[\therefore\] The equation of circle be
\[{\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0\]
As it passes through (1,−2)
\[\therefore\] Put x=1, y=−2
\[ \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0\]
\[ \Rightarrow \lambda = 4\]
\[\therefore\] equation of circle is \[{\left( {x - 3} \right)^2} + {y^2} - 8 = 0\]
Now, from the options \[\left( {5, - 2} \right)\] satisfies equation of circle. Answer the question to unlock