Question

The center of a circle is $\left( {2a,a - 7} \right)$. Find the values of $a$ if the circle passes through the point $\left( {11, - 9} \right)$ and has diameter $10\sqrt 2 $ units.
A) $\dfrac{3}{5}$
B) $3$
C) $5$
D) $\dfrac{5}{3}$

Answer 1

Hint: We are provided with the coordinates of the center of the circle and the point that lies on the circle. We know that the distance between the center to any point on the circle is equal to the radius. And also, the length of the diameter is twice that of radius.

Complete step-by-step answer:
Let us note down the data that is given in the question,
The center of the circle is located at $\left( {2a,a - 7} \right)$.
The point that passes through the circle is $\left( {11, - 9} \right)$.
The length of the diameter of the circle is $10\sqrt 2 $ units.
We know that the length of the diameter of the circle is twice of the length of the radius of the circle.
So, radius of the circle $ = \dfrac{{10\sqrt 2 }}{2}$
                      $ = 5\sqrt 2 $
We know that in a circle the distance between the centre of the circle and the point that lies on the circle is equal to the length of the radius.
So, let us find out the distance between centre $\left( {2a,a - 7} \right)$ and the point $\left( {11, - 9} \right)$ and equate that to the length of the radius of the circle.
We need to apply the distance formula to find out the distance between those two numbers.
According to the distance formula, the distance between $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ will be calculated as
$AB = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $
Already we have two points $\left( {2a,a - 7} \right)$ and $\left( {11, - 9} \right)$. So, substitute in the above formula to get the distance between them which is equal to radius of the circle.
Let $r$ is the radius of the circle.
\[
   \Rightarrow r = \sqrt {{{\left( {2a - 11} \right)}^2} + {{\left( {a - 7 + 9} \right)}^2}} \\
   \Rightarrow r = \sqrt {{{\left( {2a - 11} \right)}^2} + {{\left( {a + 2} \right)}^2}} \\
 \]
Now use the below basic algebraic formulae to simple the above equation.
$
  {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \\
  {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \\
 $
Then we will get the equation as:
\[
   \Rightarrow r = \sqrt {4{a^2} + 121 - 44a + {a^2} + 4 + 4a} \\
   \Rightarrow r = \sqrt {5{a^2} - 40a + 125} \\
 \]
Now, find the value of $a$ from the above equation.
\[
   \Rightarrow r = \sqrt {5{a^2} - 40a + 125} \\
   \Rightarrow 5{a^2} - 40a + 125 = {r^2} \\
 \]
Substitute the value of $r = 5\sqrt 2 $ in the above equation.
\[
   \Rightarrow 5{a^2} - 40a + 125 = {\left( {5\sqrt 2 } \right)^2} \\
   \Rightarrow 5{a^2} - 40a + 125 - 50 = 0 \\
   \Rightarrow 5{a^2} - 40a + 75 = 0 \\
 \]
Now, factorize the above quadratic equation to get the solution.
\[
   \Rightarrow 5{a^2} - 40a + 75 = 0 \\
   \Rightarrow {a^2} - 8a + 15 = 0 \\
 \]
Spit the between term to get according factors
\[
   \Rightarrow {a^2} - 3a - 5a + 15 = 0 \\
   \Rightarrow a\left( {a - 3} \right) - 5\left( {a - 3} \right) = 0 \\
   \Rightarrow \left( {a - 3} \right)\left( {a - 5} \right) = 0 \\
 \]
From the above we can conclude that,
$\left( {a - 3} \right) = 0$ or $\left( {a - 5} \right) = 0$
So, we will get $a = 3$ and $a = 5$ as a solution.
Hence the options B and C are correct.

Note: This type of problem can be solved in many ways. Sometimes it would be better to go with the option verification one by one. Because sometimes it would be very hard to simplify some equations. And it takes a lot of time to do that. In that situation we can go with option verification to save the time.