Question

The lines $2x+3y=5$ and $3x-4y=7$ are the diameter of the circle of area 154 square units then the equation of circle is?

Answer 1

Hint: Assume the coordinates of the center of the circle as (h, k) and the radius as r. Write the equation of the circle as ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$. Now, to find the values of the coordinates h and k solve the two equations given. The value of x will be considered as the value of k and the value of y will be considered as the value of h. Use the formula of the area of a circle given as Area = $\pi {{r}^{2}}$ and equate it with 154 to calculate the value of r. Use the value $\pi =\dfrac{22}{7}$.

Complete step by step solution:
Here we have been provided with equations of two lines which are the diameters of a circle of area 154 square units. We have been asked to determine the equation of the circle. To determine the equation of the circle we need to determine the coordinates of its center and the radius.

Now, let us assume the coordinates of the center of the circle is (h, k) and its radius is r so the equation of the circle is given as ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$. There are two equations $2x+3y=5$ and $3x-4y=7$ representing the diameters of the circle. We know that the diameters will intersect at the center of the circle so we have to solve these two equations to get the value of h and k. The x – coordinate obtained will be h and the y – coordinate obtained will be k. Let us solve the two equations.
$2x+3y=5$ - (1)
$3x-4y=7$ - (2)
Multiplying equation (1) with 3 and equation (2) and taking the difference we get,
$\begin{align}
  & \Rightarrow \left( 6x+9y \right)-\left( 6x-8y \right)=15-14 \\
 & \Rightarrow 17y=1 \\
 & \Rightarrow y=\dfrac{1}{17} \\
\end{align}$
Substituting the above obtained value of y in equation (1) we get,
$\begin{align}
  & \Rightarrow 2x+\dfrac{3}{17}=5 \\
 & \Rightarrow 2x=5-\dfrac{3}{17} \\
 & \Rightarrow x=\dfrac{41}{17} \\
\end{align}$
So the coordinates of the center of the circle is $\left( h,k \right)=\left( \dfrac{41}{17},\dfrac{1}{17} \right)$. Now, let us find the radius of the circle. The area of the circle is given 154 square units and we know that the formula of the area is Area = $\pi {{r}^{2}}$, so substituting the given values we get,
$\Rightarrow \pi {{r}^{2}}=154$
Using the value $\pi =\dfrac{22}{7}$ and simplifying we get,
$\begin{align}
  & \Rightarrow \dfrac{22}{7}\times {{r}^{2}}=154 \\
 & \Rightarrow {{r}^{2}}={{7}^{2}} \\
\end{align}$
Finally, substituting the values of h, k and ${{r}^{2}}$ in the equation of the circle we get,
$\therefore {{\left( x-\dfrac{1}{17} \right)}^{2}}+{{\left( y-\dfrac{41}{17} \right)}^{2}}={{7}^{2}}$
Hence the above relation is our answer.

Note: Note that here we have found the equation of the circle in its general form. You can also find the equation in the standard form given as ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ where the coordinates of the center is $\left( -g,-f \right)$ and radius is ${{r}^{2}}=\left( {{g}^{2}}+{{f}^{2}}-c \right)$. Remember the equation of a circle in both the forms. You can convert the general form into the standard form by simply expanding the obtained equation using the required algebraic identity. We have used the value $\pi =\dfrac{22}{7}$ for the ease of calculations.