Question:
If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$, then the equation of the circle is
If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$, then the equation of the circle is
Updated On: Jan 30, 2025
- $ x^2 + y^2 + 2x - 4y - 47 = 0$
- $x^2 + y^2 = 49$
- $x^2 + y^2 -2x + 2y -47 = 0$
- $x^2 + y^2 = 17$
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The Correct Option is C
Solution and Explanation
Let $\left( x _{0}, y _{0}\right)$ be the center of the circle
Intersection of any $2$ diameter gives us the center.
$x =\frac{5+3 y }{2}$ substituting this in
$3 x-4 y=7$
$3 \frac{5+3 y}{2}-4 y=7$
$\Rightarrow y+1=0$
$\Rightarrow y=-1$
$x_{0}=\frac{5+3(-1)}{2}=1$
Equation of circle will be
$\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=r^{2}$
$(x-1)^{2}+(y+1)^{2}=7^{2}$
$\Rightarrow x^{2}+y^{2}-2 x+2 y-47=0$
Intersection of any $2$ diameter gives us the center.
$x =\frac{5+3 y }{2}$ substituting this in
$3 x-4 y=7$
$3 \frac{5+3 y}{2}-4 y=7$
$\Rightarrow y+1=0$
$\Rightarrow y=-1$
$x_{0}=\frac{5+3(-1)}{2}=1$
Equation of circle will be
$\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=r^{2}$
$(x-1)^{2}+(y+1)^{2}=7^{2}$
$\Rightarrow x^{2}+y^{2}-2 x+2 y-47=0$
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