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The equation of the circle that passes through (-1,6) and which has a center at (2,3) can be written as x^2 + y^2 + Ax + By + C = 0. Find A*B*C

 

 

Please help thanks!!!

 Mar 6, 2019
 #1
avatar+6046 
+2

\(r^2 = (-1 - 2)^2 + (6-3)^2 = 18\\ (x-2)^2 + (y-3)^2 = 18\\ x^2 - 4x + 4 + y^2 - 6y + 9 = 18\\ x^2 + y^2 -4x -6y-5 = 0\\ (-4)(-6)(-5) = -120\)

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 Mar 6, 2019
 #2
avatar+23850 
+3

The equation of the circle that passes through (-1,6) and which has a center at (2,3)
can be written as x^2 + y^2 + Ax + By + C = 0.
Find A*B*C

 

\(\text{Let $x_c=2,\ y_c =3,\ x_p=-1,\ y_p=6$} \)

 

\(\begin{array}{|rcll|} \hline (x-x_c)^2+(y-y_c)^2 &=& r^2 \quad | \quad r^2 = (x_p-x_c)^2+(y_p-y_c)^2 \\ (x-x_c)^2+(y-y_c)^2 &=& (x_p-x_c)^2+(y_p-y_c)^2 \\ x^2-2xx_c+ \not{x_c^2}+y^2-2yy_c+\not{y_c^2} &=& x_p^2-2x_px_c+\not{x_c^2}+y_p^2-2y_py_c+\not{y_c^2} \\ x^2-2xx_c +y^2-2yy_c &=& x_p^2-2x_px_c +y_p^2-2y_py_c \\ x^2+y^2+(-2x_c)x+(-2y_c)y &=& x_p^2-2x_px_c +y_p^2-2y_py_c \\ x^2+y^2+ (-2x_c) x+ (-2y_c)y -(x_p^2-2x_px_c +y_p^2-2y_py_c ) &=& 0 \\ x^2+y^2+\underbrace{(-2x_c)}_{=A}x+\underbrace{(-2y_c)}_{=B}y +\underbrace{(2x_px_c+2y_py_c -x_p^2-y_p^2)}_{=C} &=& 0 \\\\ A&=& -2x_c\\ &=& -2\cdot 2 \\ \mathbf{A} &=& \mathbf{-4} \\\\ B&=& -2y_c\\ &=& -2\cdot 3 \\ \mathbf{B} &=& \mathbf{-6} \\\\ C &=& 2x_px_c+2y_py_c -x_p^2-y_p^2 \\ &=& 2(-1)2+2\cdot 6 \cdot 3-(-1)^2-(6)^2 \\ &=& -4+36-1-36 \\ \mathbf{C} &=& \mathbf{-5} \\\\ A\cdot B\ \cdot C &=& (-4)(-6)(-5) \\ &=& -4\cdot 5 \cdot 6 \\ \mathbf{A\cdot B \cdot C } &=& \mathbf{-120} \\ \hline \end{array} \)

 

 

laugh

 Mar 6, 2019

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