Question

The diameter of a circle is along $2x+y-7=0$ and$x+3y-11=0$. Then the equation of this circle which also passes through $\left( 5,7 \right)$ is
A) ${{x}^{2}}+{{y}^{2}}-4x-6y-16=0$
B) ${{x}^{2}}+{{y}^{2}}-4x-6y-20=0$
C) ${{x}^{2}}+{{y}^{2}}-4x-6y-12=0$
D) ${{x}^{2}}+{{y}^{2}}+4x+6y-12=0$

Answer 1

Here we have to calculate the equation of the circle. For that, we will first find the point of centre of the circle and length of radius of the circle. We will find the point of the centre of the circle by solving the two equations of the diameter given by the elimination method as the intersection of two diameters give the center of the circle. From there, we will get the point of centre. Then for radius, we will calculate the distance between the point on the center and the point on the circumference of the circle by using the distance formula. From there we will find the equation of the circle using the point of center of the circle and length of the radius of the circle.

Complete step by step solution:
The given equations of two diameters are:-
$2x+y-7=0$…………….$\left( 1 \right)$
$x+3y-11=0$…………… $\left( 2 \right)$
Now, we will calculate the center of the circle by solving these two equations of diameter as we know the intersection of two diameters give the center of the circle.
Multiplying equation $\left( 2 \right)$by 2 and subtracting it from equation$\left( 1 \right)$, we get
$\Rightarrow 2x+y-7-2\left( x+3y-11 \right)=0$
Simplifying it further, we get
$\Rightarrow y-7-6y+22=0 \\\ \Rightarrow 5y=15 \\\$
Now, we will divide 15 by 5.
$\Rightarrow y=3$
We will put the value of y in equation $\left( 2 \right)$
$\Rightarrow x+3\times 3-11=0 \\\ \Rightarrow x=2 \\\$
Therefore, the point of center of the circle is $\left( 2,3 \right)$
It is given that the circle passes through the point $\left( 5,7 \right)$. The distance between this point and the point of center will give the radius.
Now, we will calculate the distance between the point $\left( 5,7 \right)$ and $\left( 2,3 \right)$ using the distance formula.
Distance of the point $(2,3)$ from $(5,7)$ $=\sqrt{\left[ {{\left( 5-2 \right)}^{2}}+{{\left( 7-3 \right)}^{2}} \right]}$
We will calculate the difference of the terms inside the bracket.
Distance of the point $(2,3)$ from $(5,7)$ $=\sqrt{\left[ {{3}^{2}}+{{4}^{2}} \right]}$
Now, we will calculate the square of these two terms and then we will add both the terms.
Distance of the point $(2,3)$ from $(5,7)$ $=\sqrt{25}$
The square root of 25 is 5.
Distance of the point $(2,3)$ from $(5,7)$ $=5$
We know, the equation of the circle is
$\Rightarrow {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$
$\Rightarrow \left( h,k \right)=\text{center of circle}$
$r=radius$
Putting the value of center and radius in the equation of the circle, we get.
$\Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{5}^{2}}$
Applying exponents on the bases, we get
$\Rightarrow {{x}^{2}}+4-4x+{{y}^{2}}+9-6y=25$
Simplifying the equation, we get
$\Rightarrow {{x}^{2}}+{{y}^{2}}-4x-6y-12=0$

Therefore, the required equation of circle is
${{x}^{2}}+{{y}^{2}}-4x-6y-12=0$
Thus, the correct option is C.

Note:
The formula which we have used here for finding the distance between the two points is in two dimensional.
The distance formula for three dimensional is different.
Say we have to find the distance between the points $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and $({{x}_{2}},{{y}_{2}},{{z}_{2}})$
So the distance between these two points is
$\sqrt{\left[ {{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}} \right]}$