Question

Find the equation of the circle which passes through $\left( {3, - 2} \right)$, $\left( { - 2,0} \right)$ and has its center on the line $2x - y = 3$.

Answer 1

Hint: In this question, we are given the points through which the circle is passing. Also, the equation of the line passing through the center. Using the general form of a circle whose center is at $\left( { - g, - f} \right)$. Convert $2x - y = 3$ using the point $\left( { - g, - f} \right)$. Put the passing point in the equation one by one. You’ll get two equations to solve them. Lastly, put the values of $f,g,c$in the equation of the circle.

Formula Used:
The general form of the equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$whose center is $\left( { - g, - f} \right)$

Complete step by step solution:
Given that,
The circle is passing through $\left( {3, - 2} \right)$, $\left( { - 2,0} \right)$and center is on the line $2x - y = 3$
As we know that, general equation of the circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0 - - - - (1)$whose center is at $\left( { - g, - f} \right)$
Therefore, the line passing through the center will be $ - 2g + f = 3$ OR $f = 3 + 2g - - - - \left( 2 \right)$
Now, the circle is passing through $\left( {3, - 2} \right)$, $\left( { - 2,0} \right)$
Using equation (1)
Equation of circle at $\left( {3, - 2} \right)$,
${3^2} + {\left( { - 2} \right)^2} + 2g\left( 3 \right) + 2f\left( { - 2} \right) + c = 0$
$9 + 4 + 6g - 4f + c = 0$
From equation (2)
$13 + 6g - 4\left( {3 + 2g} \right) + c = 0$
$13 + 6g - 12 - 8g + c = 0$
$ - 2g + c = - 1$
$2g - c = 1$
Multiply above equation by $2$
$4g - 2c = 2 - - - - - \left( 3 \right)$
Now, equation of circle at $\left( { - 2,0} \right)$
${\left( { - 2} \right)^2} + {\left( 0 \right)^2} + 2g\left( { - 2} \right) + 2f\left( 0 \right) + c = 0$
$4 - 4g + c = 0$
$4g - c = 4 - - - - - \left( 4 \right)$
On solving (3) and (4),
We get $g = \dfrac{3}{2}$, $c = 2$
From equation (2), $f = 6$
Hence, using equation (1). The equation of the circle is passes through $\left( {3, - 2} \right)$, $\left( { - 2,0} \right)$ and has its center on the line $2x - y = 3$is ${x^2} + {y^2} + 3x + 12y + 2 = 0$

Note: A circle is a closed curve drawn from a fixed point known as the center, with all points on the curve being the same distance from the center point. While solving such questions always keep in mind to find the equation we need the value of the radius and the point of center. Also, one should always remember each and every equation of circle (General and standard both). Also, the formulas to find the radius and the distance between any point and line.