We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
115
1
avatar+1229 

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
 #1
avatar+23137 
+2

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 \(\left(\overline{AB}\right)^2\).

 

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

 

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

 

 

 

laugh

 Jul 22, 2019

39 Online Users

avatar
avatar