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# help

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The circle centered at $$(2,-1)$$ and with radius $$4$$ intersects the circle centered at $$(2,5)$$ and with radius $$\sqrt{10}$$ at two points $$A$$ and $$B$$. Find $$(AB)^2$$.

Jul 21, 2019

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The circle centered at $$(2,-1)$$ and with radius $$4$$ intersects the circle centered at $$(2,5)$$ and with radius $$\sqrt{10}$$ at two points $$A$$ and $$B$$.
Find $$\left(\overline{AB}\right)^2$$.

$$\begin{array}{|lrcll|} \hline (1) & (x-2)^2 +(y-5)^2 &=& 10 \\ & (x-2)^2 +(y-(-1))^2 &=& 4^2 \\ (2) & (x-2)^2 +(y+1)^2 &=& 16 \\ \hline (2)-(1): & (x-2)^2 +(y+1)^2 -((x-2)^2 +(y-5)^2) &=& 16-10 \\ & (x-2)^2 +(y+1)^2 -(x-2)^2 -(y-5)^2 &=& 6 \\ & (y+1)^2 -(y-5)^2 &=& 6 \\ & y^2+2y+1-(y^2-10y+25) &=& 6 \\ & y^2+2y+1-y^2+10y-25 &=& 6 \\ & 12y+1-25 &=& 6 \\ & 12y-24 &=& 6 \quad &| \quad : 6 \\ & 2y-4 &=& 1 \\ & 2y &=& 1+4 \\ & 2y &=& 5 \\ & \mathbf{ y } &=& \mathbf{\dfrac{5}{2}} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1) & (x-2)^2 +(y-5)^2 &=& 10 \quad | \quad y=\dfrac{5}{2} \\ & (x-2)^2 +\left(\dfrac{5}{2}-5\right)^2 &=& 10 \\ & (x-2)^2 +\left(-\dfrac{5}{2}\right)^2 &=& 10 \\ & (x-2)^2 + \dfrac{25}{4} &=& 10 \\ & (x-2)^2 &=& 10 - \dfrac{25}{4} \\ & (x-2)^2 &=& \dfrac{15}{4} \\ & x-2 &=& \pm\dfrac{\sqrt{15}}{2} \\ & x &=& 2 \pm \dfrac{\sqrt{15}}{2} \\ \\ & \mathbf{x_1} &=& 2 + \dfrac{\sqrt{15}}{2} \\ & \mathbf{x_2} &=& 2 - \dfrac{\sqrt{15}}{2} \\\\ & \mathbf{ \left(\overline{AB}\right)^2} &=& (x_1-x_2)^2 \\ & &=& \left(2 + \dfrac{\sqrt{15}}{2}- \left(2 - \dfrac{\sqrt{15}}{2}\right) \right)^2 \\ & &=& \left(2 + \dfrac{\sqrt{15}}{2}-2 + \dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(\dfrac{\sqrt{15}}{2}+ \dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(2\dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(\sqrt{15}\right)^2 \\ & &=& \mathbf{15} \\ \hline \end{array}$$

Jul 22, 2019