1 problem found
Show that the equation of any circle passing through the points of intersection of the ellipse \((x+2)^2 +2y^2 =18\) and the ellipse \(9(x-1)^2 +16y^2 = 25\) can be written in the form \[ x^2-2ax +y^2 =5-4a\;. \]
Solution: \begin{align*} && (x+2)^2 +2y^2 &=18 \\ && 9(x-1)^2 +16y^2 &= 25 \\ \Rightarrow && 2y^2 &= 18 - (x+2)^2 \\ && 16y^2 &= 25 - 9(x-1)^2 \\ \Rightarrow && 25-9(x-1)^2 &= 8 \cdot 18 - 8(x+2)^2 \\ \Rightarrow && 25 -9+18x-9x^2 &= 144 -32- 32x +8x^2 \\ \Rightarrow && 0 &= 96 - 50x+x^2 \\ &&&= (x-48)(x-2) \\ \Rightarrow && x &= 2,48 \\ \Rightarrow && 2y^2 &= 2, 18-50^2 \\ \Rightarrow && (x,y) &= (2,\pm1) \end{align*} Therefore any circle must have it's centre on there perpendicular bisector of \((2, \pm 1)\), ie on the \(x\)-axis. Therefore it will have equation \((x-a)^2+y^2 = r^2\) and also contain the point \((2,1)\), therefore: \begin{align*} r^2 &= (2-a)^2 + 1^2 \\ &= 4 -2a+a^2 + 1 \\ &= 5-2a+a^2 \end{align*} and the equation is: \begin{align*} && (x-a)^2 + y^2 &= 5-4a+a^2 \\ \Rightarrow && x^2-2ax+a^2 +y^2 &= 5-4a+a^2 \\ \Rightarrow && x^2-2ax+y^2 &= 5-4a \end{align*} as required.