The equation of the circle shown in the following diagram can be written as $x^2 + Ay^2 + Bx + Cy + D = 0$. Find $A+B+C+D$.
+9481 the center of the circle = (-1, 1)
the radius of the circle = \(\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5\)
So the equation of the circle is...
\((x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0 \)
A + B + C + D = 1 + 2 - 2 - 3 = -2
Here's a graph to check the equation of the circle:
+9481 the center of the circle = (-1, 1)
the radius of the circle = \(\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5\)
So the equation of the circle is...
\((x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0 \)
A + B + C + D = 1 + 2 - 2 - 3 = -2
Here's a graph to check the equation of the circle:
