Question

Find the intersection of the circle x^2+ y^2 = 29 and y = x - 3 algebraically

Answer 1

Answer:

Intersections: (-2, -5) and (5, 2)

Hope this helps!

Answer 2

$x^{2} +y^{2} =29$ circle and $y=x-3$

$y^{2} =29-x^{2}$                   $\left ( y \right )^{2}=\left ( x-3 \right )^{2}$

                                       $y^{2}=x^{2}-6x+9$

$\Rightarrow x^{2} -6x+9=29-x^{2}$  

$\Rightarrow 2x^{2} -6x-20=0$         $\Rightarrow 2x^{2} -3x-10=0$

$x^{2} -5x+2x-10=0$       $\Rightarrow x\left ( x-5 \right ) +2\left ( x-5 \right )=0$

$x=5, \ x=-2$

$\Rightarrow y=29-25=4\pm 2$       $\Rightarrow y=29-4=25\pm 5$

The circle intersects the line at $\left ( 5,2 \right )$ and $\left ( -2,-5 \right )$.