Question

Find the coordinates of the centres and the radii of the circles whose equations are: (i) $2x^2 + 2y^2 = 3x - 5y + 7$ (ii) $x^2 + y^2 - 4x - 8y = 41$ (iii) $\sqrt{1+ m^2} (x^2 + y^2) - 2cx - 2mcy = 0$

Answer 1

(i) Center: $\left(\frac{3}{4}, -\frac{5}{4}\right)$, Radius: $\frac{3\sqrt{10}}{4}$ (ii) Center: $(2, 4)$, Radius: $\sqrt{61}$ (iii) Center: $\left(\frac{c}{\sqrt{1+ m^2}}, \frac{mc}{\sqrt{1+ m^2}}\right)$, Radius: $|c|$

Answer 2

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, for which the center is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$.

To solve, each given equation is transformed into the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$. This is achieved by ensuring the coefficients of $x^2$ and $y^2$ are 1, and then rearranging terms. The values of $g$, $f$, and $c$ are identified to calculate the center $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.

(i) $2x^2 + 2y^2 = 3x - 5y + 7$ Divide by 2: $x^2 + y^2 = \frac{3}{2}x - \frac{5}{2}y + \frac{7}{2}$ Rearrange to standard form: $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$ Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$: $2g = -\frac{3}{2} \implies g = -\frac{3}{4}$ $2f = \frac{5}{2} \implies f = \frac{5}{4}$ $c = -\frac{7}{2}$ Center: $(-g, -f) = \left(\frac{3}{4}, -\frac{5}{4}\right)$ Radius: $r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3}{4}\right)^2 + \left(\frac{5}{4}\right)^2 - \left(-\frac{7}{2}\right)}$ $r = \sqrt{\frac{9}{16} + \frac{25}{16} + \frac{56}{16}} = \sqrt{\frac{90}{16}} = \frac{3\sqrt{10}}{4}$

(ii) $x^2 + y^2 - 4x - 8y = 41$ Rearrange to standard form: $x^2 + y^2 - 4x - 8y - 41 = 0$ Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$: $2g = -4 \implies g = -2$ $2f = -8 \implies f = -4$ $c = -41$ Center: $(-g, -f) = (2, 4)$ Radius: $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-4)^2 - (-41)}$ $r = \sqrt{4 + 16 + 41} = \sqrt{61}$

(iii) $\sqrt{1+ m^2} (x^2 + y^2) - 2cx - 2mcy = 0$ Divide by $\sqrt{1+ m^2}$: $x^2 + y^2 - \frac{2c}{\sqrt{1+ m^2}}x - \frac{2mc}{\sqrt{1+ m^2}}y = 0$ Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$: $2g = -\frac{2c}{\sqrt{1+ m^2}} \implies g = -\frac{c}{\sqrt{1+ m^2}}$ $2f = -\frac{2mc}{\sqrt{1+ m^2}} \implies f = -\frac{mc}{\sqrt{1+ m^2}}$ $c = 0$ Center: $(-g, -f) = \left(\frac{c}{\sqrt{1+ m^2}}, \frac{mc}{\sqrt{1+ m^2}}\right)$ Radius: $r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{c}{\sqrt{1+ m^2}}\right)^2 + \left(-\frac{mc}{\sqrt{1+ m^2}}\right)^2 - 0}$ $r = \sqrt{\frac{c^2}{1+ m^2} + \frac{m^2c^2}{1+ m^2}} = \sqrt{\frac{c^2(1+m^2)}{1+m^2}} = \sqrt{c^2} = |c|$