Question

The centre of a circle passing through the point (0,0), (1,0) and touching the circle $x^2 + y^2 = 9$ is?
A. $(\dfrac{3}{2}, \dfrac{1}{2})$
B.  $(\dfrac{1}{2}, \dfrac{3}{2})$
C.  $(\dfrac{1}{2}, \dfrac{1}{2})$
D. None of these

Answer 1

Hint: Here we will use the basic equation of circle for finding the centre and put the given values in the equation.

Complete step by step solution:
Given: The circle passes through (0,0), (1,0) and touches the circle $x^2 + y^2 = 9$.

Let the centre of the circle C be (a, b) and radius r
Circle \[{x^2} + {\rm{ }}{y^2} = {\rm{ }}9\] has centre \[\left( {0,0} \right){\rm{ }}\] and radius is $3$.
Therefore, the basic equation of a circle C having centre (a, b) and radius $r$ is \[{\left( {x - a} \right)^2} + {\rm{ }}{\left( {y - b} \right)^2} = {\rm{ }}{r^2}\]
According to question the circle passes through (0,0)
Putting (0,0) in the equation
\[{\left( {0 - a} \right)^2} + {\rm{ }}{\left( {0 - b} \right)^2} = {\rm{ }}{r^2}\]
\[{a^2} + {\rm{ }}{b^2} = {\rm{ }}{r^2}\] ………. (i)
Also, given the circle passes through (1,0)
Putting in the equation
\[{\left( {1 - a} \right)^2} + {\rm{ }}{\left( {0 - b} \right)^2} = {\rm{ }}{r^2}\]
\[{\left( {1 - a} \right)^2} + {\rm{ }}{\left( b \right)^2} = {\rm{ }}{r^2}\] ………. (ii)
Subtracting equation (ii) from equation (i)
\[\Rightarrow {\left( {1 - a} \right)^2} + {\rm{ }}{\left( b \right)^2}-{\rm{ }}\left( {{a^2} + {\rm{ }}{b^2}} \right){\rm{ }} = {\rm{ }}{r^2} - {\rm{ }}{r^2}\]

$\Rightarrow (1-a)^2 - a^2 =0$

$\Rightarrow (1-a)^2 =a^2$

$\Rightarrow 1+a^2-2a=a^2$

$\Rightarrow 1-2a=0$

$\Rightarrow 2a=1$

$\Rightarrow a= \dfrac{1}{2}$

Since it is given that circle C is touching the circle \[{x^2} + {\rm{ }}{y^2} = {\rm{ }}9\] and given it passes through (0,0), (1,0) this means that $a = \dfrac{1}{2}$ is inside the circle.
This means difference between the centers of the two circles is $3-r$
Let D be the difference between the centers of the two circles
\[D{\rm{ }} = {\rm{ }}3 - r\]
\[\Rightarrow 3 - r = \sqrt {({{(a - 0)}^2} + {{(b - 0)}^2})} \]
Squaring both sides
\[{\left( {3 - r} \right)^2} = {\rm{ }}{a^2} + {\rm{ }}{b^2}\]
\[\Rightarrow 9{\rm{ }} + {\rm{ }}{r^2} - {\rm{ }}6r{\rm{ }} = {\rm{ }}{a^2} + {\rm{ }}{b^2}\]
 Using equation (i)
\[9{\rm{ }} + {\rm{ }}{r^2} - {\rm{ }}6r{\rm{ }} = {\rm{ }}{r^2}\;\;\]
\[\Rightarrow 6r{\rm{ }} = {\rm{ }}9\]

$\Rightarrow r=\dfrac{3}{2}$

Putting the values of a and r in (i)

${\dfrac{1}{2}}^2+b^2={\dfrac{3}{2}}^2$
$\Rightarrow b=2$

Therefore, the centre of the circle is $(\dfrac{1}{2}, 2)$. So, Option (D) is correct.

Note: In such type of questions we firstly find the centre and radius of the circle and make the equations according to the problem statement.