Question

If the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ lie along diameters of a circle of circumference $10\pi $, then find the equation of the circle.
A. ${x^2} + {y^2} - 2x + 2y - 23 = 0$
B. ${x^2} + {y^2} + 2x - 2y - 23 = 0$
C. ${x^2} + {y^2} + 2x + 2y - 23 = 0$
D. ${x^2} + {y^2} - 2x - 2y - 23 = 0$

Answer 1

Hint: First suppose that the coordinate of the center is $(h,k)$ and r be the center. Equate the given circumference $10\pi $ of the circle with the formula of the circumference to obtain the value of r. Then substitute (h, k) in the given equations and solve them to obtain the values of h and k. Then use the formula of a circle and substitute the values of center and radius to obtain the required equation.

Formula Used:
The formula of circumference of a circle is $2\pi r$ , where r is the radius.
The equation of a circle is,
${(x - a)^2} + {(y - b)^2} = {r^2}$ , where (a, b) is the coordinate of the center and r is the radius.

Complete step by step solution:
It is given that the circumference is $10\pi $
So,
$2\pi r = 10\pi $
$ \Rightarrow r = 5$
The diameter of a circle passes through the center,
So, $2h + 3k + 1 = 0$ --(1)
$3h - k - 4 = 0$---(2)
Multiply (2) by 3 and add with (1) to obtain the value of h.
$2h + 3k + 1 + 9h - 3k - 12 = 0$
$11h - 11 = 0$
$h = 1$
Substitute $h = 1$in the equation (1) to obtain the value of k.
$2.1 + 3k + 1 = 0$
$3k + 3 = 0$
$k = - 1$
Substitute 1 for a, -1 for b and 5 for r in the equation ${(x - a)^2} + {(y - b)^2} = {r^2}$to obtain the required equation.
${(x - 1)^2} + {(y + 1)^2} = {5^2}$
${x^2} - 2x + 1 + {y^2} + 2y + 1 = 25$
${x^2} + {y^2} - 2x + 2y - 23 = 0$

Option ‘A’ is correct

Note: Whenever we face such types of questions the key concept here is to have a good understanding of the general equation of circle.