The circle passing through (1, −2) and touching the x-axis at (3,0) also passes through the point.
(A).(-5, 2)
(B). (2, -5)
(C). (5, -2)
(D). (-2, 5

Hint: For solving this question, we let the centre be (3, k) and radius of the circle be k. We know that the equation of the circle is: \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]. Putting all the values in the equation and satisfying (1, -2), by this we get the value of k. After finding the value of k, we also check the valid points.

Complete step-by-step solution -
We know the equation of the circle is:
\[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]
Let, centre be (3, k) and r is k.
Now, the required equation of the circle is:
 \[{{\left( x-3 \right)}^{2}}+{{\left( y-k \right)}^{2}}={{k}^{2}}\]
It is given that it passes through point (1, -2). So, this point satisfies the equation also,
${{\left( 1-3 \right)}^{2}}+{{\left( -2-k \right)}^{2}}={{k}^{2}}$
Using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we expand the above equation,
$\begin{align}
  & \Rightarrow 4+4+{{k}^{2}}+4k={{k}^{2}} \\
 & \Rightarrow 4k=-8 \\
 & \Rightarrow k=\dfrac{-8}{4} \\
 & \Rightarrow k=-2 \\
\end{align}$
So, the required equation of the circle is:
\[{{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=4\]

Now, we have to also check the valid point also.
Putting the point (-5, 2) in the equation,
\[\begin{align}
  & {{\left( -5-3 \right)}^{2}}+{{\left( 2+2 \right)}^{2}}=4 \\
 & {{\left( -8 \right)}^{2}}+{{\left( 4 \right)}^{2}}=4 \\
 & \Rightarrow 64+16\ne 4 \\
\end{align}\]
Left-hand side is not equal to the right-hand side.
Hence, point (-5, 2) does not satisfy the equation.
Putting the point (2, -5) in the equation,
\[\begin{align}
  & {{\left( 2-3 \right)}^{2}}+{{\left( -5+2 \right)}^{2}}=4 \\
 & {{\left( -1 \right)}^{2}}+{{\left( -3 \right)}^{2}}=4 \\
 & \Rightarrow 10\ne 4 \\
\end{align}\]
        Left-hand side is not equal to the right-hand side.
        Hence, point (2, -5) does not satisfy the equation.
Putting the point (5, -2) in the equation,
\[\begin{align}
  & {{\left( 5-3 \right)}^{2}}+{{\left( -2+2 \right)}^{2}}=4 \\
 & {{\left( 2 \right)}^{2}}+\left( 0 \right)=4 \\
 & \Rightarrow 4=4 \\
\end{align}\]
        Left hand is equal to the right-hand side.
        Hence, point (5, -2) satisfies the equation.
Putting the point (-2, 5) in the equation,
\[\begin{align}
  & {{\left( -2-3 \right)}^{2}}+{{\left( 5+2 \right)}^{2}}=4 \\
 & {{\left( -5 \right)}^{2}}+{{\left( 7 \right)}^{2}}=4 \\
 & \Rightarrow 25+49\ne 4 \\
\end{align}\]
Left-hand side is not equal to the right-hand side.
       Hence, point (-2, 5) not satisfy the equation
Therefore, option (c) is correct.

Note: Students must know the general form of circle for solving this problem. All the options must be checked individually even after obtaining the correct answer as multiple answers are possible. Calculations must be done correctly with the given coordinates.