✒️EQUATION
[tex] \small \boxed{\begin{array}{l} \textsf{Find the equation of the circle passes through} \\ \textsf{the points of intersection of the circles}\\ \sf x^2 + y^2 = 5, x^2 + y^2 - x + y = 4\textsf{ and through} \\ \textsf{the point (-2,3).}\end{array}} [/tex]
[tex]\small\begin{array}{|c|c} \hline\bold{Given:}\begin{cases} \: \textsf{Points of intersection of} \\ \: \sf x^2 + y^2 = 5; \\ \: \sf x^2 + y^2 - x + y = 4\ \ \&; \\ \: \sf (-2, 3) \end{cases} \\ \\ \bold{Required:}\ \textsf{Equation of the circle} \\ \\ \textsf{First, solve for the points of intersection of the} \\ \textsf{two given circles.} \\ \\ \begin{array}{l} \sf x^2 + y^2 = 5 & \sf (1) \\ \sf x^2 + y^2 - x + y = 4 & \sf (2)\end{array} \\ \\ \textsf{Subtracting (2) from (1), we get} \\\\ \sf x - y = 1 \implies y = x - 1 \quad (3) \\ \\ \textsf{Substitute (3) to (1) then solve for x.} \\ \\ \begin{array}{c} \sf x^2 + (x - 1)^2 = 5 \\ \sf x^2 + x^2 - 2x + 1 = 5 \\ \sf 2x^2 - 2x - 4 = 0 \\ \sf x^2 - x - 2 = 0 \\ \sf (x + 1)(x - 2) = 0 \end{array} \\ \\ \begin{array}{c|c} \sf x + 1 = 0 & \sf x - 2 = 0 \\ \sf x = - 1 & \sf x = 2 \\ \\ \sf y = x - 1 & \: \\ \\ \sf y = - 1 - 1 & \sf y = 2 - 1 \\ \sf y = -2 & \sf y = 1 \end{array} \\ \\ \textsf{Points of intersection:}\: \sf \{(-1, -2), (2, 1)\} \\\hline\end{array} [/tex]
[tex] \small\begin{array}{|c|c}\hline \textsf{Using the general form of the equation of the} \\ \quad \:\:\textsf{circle } \sf x^2 + y^2 + Ax + By + C = 0, \\ \\ \sf \underline{\textsf{At }(-1,-2)}, \\ \\ \sf (-1)^2 + (-2)^2 - A - 2B + C = 0 \\ \sf A + 2B - C = 5\quad (1) \\ \\ \sf \underline{\textsf{At }(2, 1),} \\ \\ \sf 2^2 + 1^2 + 2A + B + C = 0 \\ \sf 2A + B + C = -5 \quad(2) \\ \\ \sf \underline{\textsf{At }(-2, 3)}, \\ \\ \sf (-2)^2 + 3^2 - 2A + 3B + C = 0 \\ \sf 2A - 3B - C = 13\quad(3) \\ \\ \sf \textsf{Now, solve the simultaneous equations.} \\ \\ \sf Adding\ (1)\textsf{ and }(2),\textsf{we get} \\ \\ \begin{aligned} \sf A + 2B - C &=\sf 5 \\ \sf +\quad 2A +\:\:B + C&= \sf-5\\ \hline \sf 3A \:+\: 3B\: \quad \:\: &= \sf 0 \end{aligned} \\ \quad \:\:\:\sf A + B = 0 \implies A = - B \\ \\ \textsf{Substitute }\sf A = - B \textsf{ to (1) and (3) then solve for} \\ \textsf{B and C.}\\ \\ \begin{array}{c c} \sf A + 2B - C = 5 & \sf (1) \\ \sf -B + 2B - C = 5 & \: \sf \\ \sf B - C = 5 & \: \sf (4) \\ \\ \sf 2A - 3B - C = 5 & \sf (2)\\ \sf -2B - 3B - C = 13 & \: \\ \sf -5B - C = 13 & \: \\ \sf 5B + C = -13 & \sf (5) \end{array} \\ \\ \textsf{Adding (4) and (5) implies} \\ \\ \sf 6B = -8 \implies B = -\dfrac{4}{3}, so\ A = \dfrac{4}{3}\\ \\ \textsf{Substituting the value of B to (4), we get} \\ \\ \sf -\dfrac{4}{3} - C = 5 \implies C = -\dfrac{19}{3} \\ \\ \textsf{Plug in the values of A, B, and C to the general} \\ \textsf{equation of the circle.} \\ \\ \textsf{Therefore, the equation of the circle is} \\ \qquad\red{\boxed{\sf x^2 + y^2 + \dfrac{4}{3}x - \dfrac{4}{3}x - \dfrac{19}{3} = 0}} \\ \hline\end{array} [/tex]
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[tex]\qquad\qquad\qquad\qquad\qquad\qquad\tt{fri \: 04-01-2022} \\ \qquad\qquad\qquad\qquad\qquad\qquad\tt{1:56 \: pm}[/tex]
