Question

find the equation of the circle whose diameters are along the lines 2x-3y+12 =0 and x-4y-5=0 and whose area is 154 sq units?

Answer 1

$\textbf{Given:}$

$\text{Diameters of the circle are 2x-3y+12=0 and x-4y-5=0 }$

$\text{and area of the circle is 154 square units}$

$\textbf{To find:}$

$\text{Equation of the circle}$

$\textbf{Solution:}$

$\textit{We know that, the point of intersetion of any two}$ $\textit{diameter of a circle is its centre}$

$\text{Now, we solve}$

$2x-3y+12=0$........(1)

$x-4y-5=0$............(2)

$(1)\implies\;\,2x-3y+12=0$

$(2){\times}2\implies,2x-8y-10=0$

$\text{Subtracting,}$

$5y+22=0$

$\implies\,y=\dfrac{-22}{5}$

$\text{Put ,y=\dfrac{-22}{5} in (2)}$

$x=4(\frac{-22}{5})+5$

$x=\dfrac{-88+25}{5}$

$x=\dfrac{-63}{5}$

$\implies\textbf{Centre is (\dfrac{-63}{5},\dfrac{-22}{5}) }$

$\text{Also,}$

$\text{Area of the circle=154 square units}$

$\implies\,\pi\,r^2=154$

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

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

$\implies\,r^2=7{\times}7$

$\implies\bf\,r=7$

$\textbf{Equation of the circle is}$

$\bf(x-h)^2+(y-k)^2=r^2$

$(x-(\frac{-63}{5}))^2+(y-(\frac{-22}{5}))^2=7^2$

$\bf(x+\frac{63}{5})^2+(y+\frac{22}{5})^2=49\;\;$

$\text{which is the required equation}$

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