# Determine the equation of a circle centered at (0, 0) and through the point of cutting the line 3x + 4y = 24 with the x axis?

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Mar 11, 2018

${x}^{2} + {y}^{2} = 64$

#### Explanation:

As the equation of $x$-axis i$y = 0$, we can find the point of its intersection with the line $3 x + 4 y = 24$ with the $x$-axis is obtained by putting $y - 0$.

Then we have $3 x + 4 \times 0 = 24$ or $3 x = 24$ i.e. $x = 8$ and point of intersection is $\left(8 , 0\right)$.

As center is $\left(0 , 0\right)$ and circle passes through $\left(8 , 0\right)$, its radius is $\sqrt{{\left(8 - 0\right)}^{2} + {\left(0 - 0\right)}^{2}} = \sqrt{64} = 8$

and hence equation of circle is

${\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2} = {8}^{2}$

or ${x}^{2} + {y}^{2} = 64$

graph{(x^2+y^2-64)(3x+4y-24)=0 [-16.83, 23.17, -9.36, 10.64]}

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