If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are the diameter of a circle of area $49\pi $ square units, the equation of the circle is
A. ${x^2} + {y^2} + 2x - 2y - 62 = 0$
B. ${x^2} + {y^2} - 2x + 2y - 62 = 0$
C. ${x^2} + {y^2} - 2x + 2y - 47 = 0$
D. ${x^2} + {y^2} + 2x - 2y - 47 = 0$
VerifiedHint: We can find the coordinates of the centre of the circle by solving the equation of the diameters. Then we can find the radius of the circle from the area using the equation $A = \pi {r^2}$. Now we can find the equation of the circle by substituting the values of the centre and radius in the equation ${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$. We get the required equation by simplification.
Complete step-by-step answer:
We have the equations of two diameters. We know that all the diameters meet at the center. So we can find the coordinates of the center by solving the equations.
We have equations,
$3x - 4y - 7 = 0$… (1)
$2x - 3y - 5 = 0$… (2)
To solve we can subtract 2 times equation (1) from 3 times equation (2)
$\,\,\,\,\,\,\,\,\,6x - 9y - 15 = 0$
$\underline {\left( - \right)6x - 8y - 14 = 0}$
$\,\,\,\,\,\,\,\,0x - 1y - \,\,\,\,\,1\,\, = 0$
$ \Rightarrow y = - 1$
Substituting the value of y in equation (2), we get the value of x
$ \Rightarrow 2x - 3\left( { - 1} \right) - 5 = 0$
On simplification we get,
$ \Rightarrow 2x + 3 - 5 = 0$
On solving for x we get,
$ \Rightarrow 2x = 2$
On dividing the equation by 2 we get,
$ \Rightarrow x = 1$
Therefore, the centre of the circle is$\left( {1, - 1} \right)$.
We have the area of the circle as $49\pi $ square units. The equation of area is given by$A = \pi {r^2}$. On substituting the value of the area, we can find the radius.
$ \Rightarrow 49\pi = \pi {r^2}$
$ \Rightarrow {r^2} = 49$
Taking the square root on both sides, we get,
$ \Rightarrow r = 7$.
Now we have a radius of 7 units.
We know that equation of a circle with centre\[\left( {{x_0},{y_0}} \right)\]and radius r is given by${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$
On substituting the values of centre and radius, we get,
${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$
On expanding the squares we get,
$ \Rightarrow $${x^2} - 2x + 1 + {y^2} + 2y + 1 = 49$
On rearranging, we get,
$ \Rightarrow $${x^2} + {y^2} - 2x + 2y - 47 = 0$
So the required equation of the circle is ${x^2} + {y^2} - 2x + 2y - 47 = 0$
Therefore the correct answer is option C.
Note: The point of intersection of 2 lines can be found out by solving the equations of the line. A circle is the collection of all points in a plane that are equidistant from a fixed point. This fixed point is called the center of the circle and the constant distance is called the radius. A line connecting two points on a circle is called a chord. The largest chord of a circle is its diameter. It passes through the center and has a length of twice the radius.
