36=x^2+y^2−6x−8y+25

16=x^2+y^2−18x−16y+145

asked by guest
on Nov 17, 2024 at 2:22 am



You asked:

Solve the system of equations:\[\begin{aligned}36 &= {x}^{2} + {y}^{2} - 6 x - 8 y + 25\\16 &= {x}^{2} + {y}^{2} - 18 x - 16 y + 145\end{aligned}\]

MathBot Answer:

The system of equations has \(2\) solutions.

\[x = \frac{69}{13}, y = \frac{124}{13}\]\[x = 9, y = 4\]

Solve \(36 = x^{2} - 6 x + y^{2} - 8 y + 25\) for \(x\). \[x = 3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}, x = \sqrt{- \left(y - 10\right) \left(y + 2\right)} + 3\]Substitute \(3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}\) for \(x\) in \(16 = x^{2} - 18 x + y^{2} - 16 y + 145\) and simplify. $$\begin{aligned}16 &amp= x^{2} - 18 x + y^{2} - 16 y + 145 \\ 16 &= \left(3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}\right)^{2} - 18 \left(3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}\right) + y^{2} - 16 y + 145 \\ - 8 y + 12 \sqrt{- y^{2} + 8 y + 20} + 120 &= 16 \end{aligned}$$Substitute \(3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}\) for \(x\) in \(16 = x^{2} - 18 x + y^{2} - 16 y + 145\) and simplify. $$\begin{aligned}16 &amp= x^{2} - 18 x + y^{2} - 16 y + 145 \\ 16 &= \left(3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}\right)^{2} - 18 \left(3 - \sqrt{- \left(y - 10\right) \left(y + 2\right)}\right) + y^{2} - 16 y + 145 \\ - 8 y + 12 \sqrt{- y^{2} + 8 y + 20} + 120 &= 16 \end{aligned}$$Substitute \(\sqrt{- \left(y - 10\right) \left(y + 2\right)} + 3\) for \(x\) in \(16 = x^{2} - 18 x + y^{2} - 16 y + 145\) and simplify. $$\begin{aligned}16 &amp= x^{2} - 18 x + y^{2} - 16 y + 145 \\ 16 &= \left(\sqrt{- \left(y - 10\right) \left(y + 2\right)} + 3\right)^{2} - 18 \left(\sqrt{- \left(y - 10\right) \left(y + 2\right)} + 3\right) + y^{2} - 16 y + 145 \\ 8 y + 12 \sqrt{- y^{2} + 8 y + 20} - 120 &= -16 \end{aligned}$$Substitute \(4\) into \(36 = x^{2} - 6 x + y^{2} - 8 y + 25\) to solve for \(x\). $$\begin{aligned}36 &= x^{2} - 6 x - 32 + 4^{2} + 25 \\ 36 &= x^{2} - 6 x + 9 \\- x^{2} + 6 x + 27 &= 0 \\ - \left(x - 9\right) \left(x + 3\right) &= 0 \\ x = -3&, x = 9\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 9,\,y = 4\end{aligned}$$Substitute \(\frac{124}{13}\) into \(36 = x^{2} - 6 x + y^{2} - 8 y + 25\) to solve for \(x\). $$\begin{aligned}36 &= x^{2} - 6 x - \frac{992}{13} + 25 + \left(\frac{124}{13}\right)^{2} \\ 36 &= x^{2} - 6 x + \frac{6705}{169} \\- x^{2} + 6 x - \frac{621}{169} &= 0 \\ - \frac{\left(13 x - 69\right) \left(13 x - 9\right)}{169} &= 0 \\ x = \frac{9}{13}&, x = \frac{69}{13}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = \frac{69}{13},\,y = \frac{124}{13}\end{aligned}$$

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