Question

The ends of the diameter are $ \left( 6,5 \right) $ and $ \left( -12,-5 \right) $ . How do you find the equation of this circle?

Answer 1

Hint: We find the coordinates of the centre from end points of the diameter of $ \left( 6,5 \right) $ and $ \left( -12,-5 \right) $ . The distance between the centre and one end point of the diameter gives the length of the radius. From the general equation of circle \[{{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}\], we find the required equation.

Complete step by step answer:
We know that the middle point of the line of the diameter is the centre of a circle.
We have the end points of the diameter as $ \left( 6,5 \right) $ and $ \left( -12,-5 \right) $ .
The middle point of the points $ \left( 6,5 \right) $ and $ \left( -12,-5 \right) $ will be the centre.
We know that for points $ \left( a,b \right) $ and $ \left( c,d \right) $ the middle point will be $ \left( \dfrac{a+c}{2},\dfrac{b+d}{2} \right) $ .
So, the centre of our required circle is $ \left( \dfrac{6-12}{2},\dfrac{5-5}{2} \right)\equiv \left( -3,0 \right) $ .
We also know that the distance from the centre to one endpoint of the diameter is the length of the radius.
Therefore, the distance between $ \left( -3,0 \right) $ and $ \left( 6,5 \right) $ will be the radius’s length. Let’s assume the length is r.
For points $ \left( a,b \right) $ and $ \left( c,d \right) $ , the distance formula is \[\sqrt{{{\left( c-a \right)}^{2}}+{{\left( d-b \right)}^{2}}}\].
For points $ \left( -3,0 \right) $ and $ \left( 6,5 \right) $ , the distance is \[\sqrt{{{\left( 6+3 \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}=\sqrt{81+25}=\sqrt{106}\].
This means \[r=\sqrt{106}\].
Now for a centre at $ \left( \alpha ,\beta \right) $ and the radius r the equation of a circle is \[{{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}\].
Our required circle with centre at $ \left( -3,0 \right) $ and the radius \[r=\sqrt{106}\] will be
\[\begin{align}
  & {{\left( x+3 \right)}^{2}}+{{\left( y \right)}^{2}}={{\left( \sqrt{106} \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+6x+{{y}^{2}}=97 \\
\end{align}\]
The equation of the circle is \[{{x}^{2}}+6x+{{y}^{2}}=97\].

Note:
 We also can use the direct formula of finding circle’s equation from end points of the diameter $ \left( a,b \right) $ and $ \left( c,d \right) $ where the equation is $ \left( x-a \right)\left( x-c \right)+\left( y-b \right)\left( y-d \right)=0 $ .
For our given points $ \left( 6,5 \right) $ and $ \left( -12,-5 \right) $ , the circle’s equation is
 $ \begin{align}
  & \left( x-6 \right)\left( x+12 \right)+\left( y-5 \right)\left( y+5 \right)=0 \\
 & \Rightarrow {{x}^{2}}+6x-72+{{y}^{2}}-25=0 \\
 & \Rightarrow {{x}^{2}}+6x+{{y}^{2}}=97 \\
\end{align} $