Search any question & find its solution
Question:
Answered & Verified by Expert
What will be the equation of circle whose centre is $(1,2)$ and touches $X$ - axis?
Options:
Solution:
2822 Upvotes
Verified Answer
The correct answer is:
$x^{2}+y^{2}-2 x-4 y+1=0$
Given, centre $(h, k)=(1,2)$ and circle touches on $X$ axis.
$\therefore$ Radius $(r)=y$ coordinate of centre $=2$
So, equation of circle is
$$
\begin{aligned}
(x-1)^{2}+(y-2)^{2}=2^{2} \quad & {\left[\because(x-h)^{2}+(y-k)^{2}=r^{2}\right] } \\
\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4 &=4 \\
\Rightarrow \quad & {\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] } \\
x^{2}+y^{2}-2 x-4 y+1 &=0
\end{aligned}
$$
$\therefore$ Radius $(r)=y$ coordinate of centre $=2$
So, equation of circle is
$$
\begin{aligned}
(x-1)^{2}+(y-2)^{2}=2^{2} \quad & {\left[\because(x-h)^{2}+(y-k)^{2}=r^{2}\right] } \\
\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4 &=4 \\
\Rightarrow \quad & {\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] } \\
x^{2}+y^{2}-2 x-4 y+1 &=0
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.