Question

Solve my question math
​

Answer 1

$\Huge(x-1)^{2}+(y+1)^{2}=7^{2}$

$\huge\textbf{Question}$

If $\textrm{ 2x-3y=5 and 3x-4y=7 }$ are the diameters of the circle of area 154 units², find the equation of the circle.

$\huge\textbf{Explanation}$

$\begin{cases} & 2x-3y=5\\ & 3x-4y=7 \end{cases}$

Where the diameters cross each other is the center of the circle. The two lines intersect at the center, $(1,-1)$.

$\large\textbf{-Equation of circles}$

We derive this equation by squaring both sides of the distance formula.

$\cdots\longrightarrow\boxed{(x-a)^{2}+(y-b)^{2}=r^{2}}$

$\bigstar\textbf{Points to note}\bigstar$

• $(a,b)$ is the center of the circle.
• $r$ is the radius of the circle. $(r>0)$

The area of the circle is given by the formula of $\pi r^{2}$, where $\pi\approx\dfrac{22}{7}$.

By taking the approximation, we get the following equation.

$\dfrac{22}{7}\times r^{2}=154$

$r^{2}=154\times\dfrac{7}{22}$

$r^{2}=7^{2},\ \boxed{r>0}$

$\therefore r=7$

Hence, the equation of the circle is as follows.

$\cdots\longrightarrow\boxed{(x-1)^{2}+(y+1)^{2}=7^{2}}$