Question

Find the equation of a circle passing through the points $(3, - 2),\,\,(2,0)$ and hence its centre lies on the line $2x - y = 3$?

Answer 1

Hint: Here we have to find the equation of a circle which passes through the points $(3, - 2),\,\,(2,0)$ and its centre lies on the line $2x - y = 3$. To find the equation of a circle we will use the standard or general equation of a circle which is given as ${(x - h)^2} + {(y - k)^2} = {r^2}$. In the above equation $r$ is the radius of a circle, $(h,k)$ are coordinates of its centre.

Complete step by step answer:
In the given question we have to find the equation of a circle which passes through the points $(3, - 2),\,\,(2,0)$ and its centre lies on the line $2x - y = 3$. We know that the equation of a circle with centre $(h,k)$ is,
${(x - h)^2} + {(y - k)^2} = {r^2}$
Since the circle passes through$(3, - 2)$. We have
$ \Rightarrow {(3 - h)^2} + {( - 2 - k)^2} = {r^2}$

Use the identity ${(a - b)^2} = {a^2} + {b^2} = 2ab$ to solve the above equation.
$ \Rightarrow 9 + {h^2} - 6h + 4 + {k^2} + 4k = {r^2}$
$ \Rightarrow {h^2} + {k^2} - 6h + 4k + 13 = {r^2} \ldots \ldots (1)$
Since the circle passes through $( - 2,0)$. We have
${( - 2 - h)^2} + {(0 - k)^2} = {r^2}$
Use the identity ${(a - b)^2} = {a^2} + {b^2} = 2ab$ to solve the above equation.
$ \Rightarrow 4 + {h^2} + 4h + {k^2} = {r^2}$
$ \Rightarrow {h^2} + {k^2} + 4h + 4 = {r^2} \ldots \ldots (2)$

Subtracting equation $(1)$ from equation $(2)$. We get,
$ \Rightarrow {h^2} + {k^2} + 4h + 4 - {h^2} - {k^2} + 6h - 4k - 13 = {r^2} - {r^2}$
Simplifying the above equation. We get,
$ \Rightarrow 10h - 4k = 9 \ldots \ldots (3)$
Since centre $(h,k)$ lies on the line $2x - y = 3$
So, we can write it as $2h - k = 3 \ldots \ldots (4)$
Now we will solve equation $(3)$ and $(4)$ by elimination to find the value of $(h,k)$
We have
$10h - 4k = 9 \\
\Rightarrow 2h - k = 3 \\ $

Multiplying the equation $2h - k = 3$ by $( - 4)$. We have,
$10h - 4k = 9 \\
\Rightarrow - 8h + 4k = - 12 \\ $
On adding both the equations. We get,
$ \Rightarrow 2h = - 3$
$ \Rightarrow h = - \dfrac{3}{2}$
Putting the value of $h = - \dfrac{3}{2}$ in equation $(4)$. We get,
$ \Rightarrow 2 \times \left( {\dfrac{{ - 3}}{2}} \right) - k = 3$
$ \Rightarrow - 3 - k = 3$
$ \Rightarrow k = - 6$

Now, let us find the radius of a circle by putting the value $h = - \dfrac{3}{2}$ and $k = - 6$ in equation $(1)$. We get,
$ \Rightarrow {\left( {\dfrac{{ - 3}}{2}} \right)^2} + {( - 6)^2} - 6\left( {\dfrac{{ - 3}}{2}} \right) + 4( - 6) + 13 = {r^2}$
$ \Rightarrow \left( {\dfrac{9}{4}} \right) + 36 + 9 - 24 + 13 = {r^2}$
$ \Rightarrow \left( {\dfrac{9}{4}} \right) + 34 = {r^2}$
$ \Rightarrow \dfrac{{9 + 136}}{4} = {r^2}$
$ \Rightarrow {r^2} = \dfrac{{145}}{4}$
Put the value of $(h,k)$ and ${r^2}$ in the standard equation of the circle. We get,
$ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {(y + 6)^2} = \dfrac{{145}}{4}$
$ \therefore {(2x + 3)^2} + 4{(y + 6)^2} = 145$

Hence, the equation of a circle which passes through the points $(3, - 2),\,\,(2,0)$ and whose centre lie on the line $2x - y = 3$is ${(2x + 3)^2} + 4{(y + 6)^2} = 145$.

Note: We can solve these types of problems by applying the formula of distance between two points. If $({x_1},{y_1})$ and $({x_2},{y_2})$ are the two points, then the distance between these points is given by the formula $d = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} $.