8 The equation of a circle is \( x^{2}+y^{2}-4 x-6 y=12 \) (a) Find the coordinates of the centre of the circle and the radius of the circle. \[ \left(x^{2}-4 x+4\right)+\left(y^{2}-6 y+9\right)=(x-2)^{2}+(y-3)^{2}=25 \] The line \( y+2 x=5 \) intersects the circle at \( A \) and \( B \). (b) Show that \( A B \) is not the diameter of the circle.

(a) To find the center and radius of the circle, we need to complete the square for both \( x \) and \( y \) terms. Given the equation of the circle: \[ x^{2}+y^{2}-4 x-6 y=12 \] We can rewrite it as: \[ x^{2}-4 x+4+y^{2}-6 y+9=12+4+9 \] \[ (x-2)^{2}+(y-3)^{2}=25 \] Now we can see that the center of the circle is at the point \( (2, 3) \) and the radius is \( \sqrt{25} = 5 \). (b) To show that \( AB \) is not the diameter of the circle, we need to show that the distance between points \( A \) and \( B \) is not equal to the diameter of the circle. The line \( y+2 x=5 \) intersects the circle at points \( A \) and \( B \). We can find the coordinates of these points by solving the system of equations formed by the circle's equation and the line's equation. Let's solve the system: \[ (x-2)^{2}+(y-3)^{2}=25 \] \[ y+2 x=5 \] Substitute \( y \) from the second equation into the first equation: \[ (x-2)^{2}+((5-2 x)-3)^{2}=25 \] \[ (x-2)^{2}+(2-2 x)^{2}=25 \] \[ (x-2)^{2}+(4-8 x+4 x^{2})=25 \] \[ x^{2}-4 x+4+4 x^{2}-8 x+4=25 \] \[ 5 x^{2}-12 x+8=25 \] \[ 5 x^{2}-12 x-17=0 \] Now we can solve for \( x \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \] \[ x = \frac{-(-12) \pm \sqrt{(-12)^{2}-4(5)(-17)}}{2(5)} \] \[ x = \frac{12 \pm \sqrt{144+340}}{10} \] \[ x = \frac{12 \pm \sqrt{484}}{10} \] \[ x = \frac{12 \pm 22}{10} \] This gives us two possible values for \( x \): \[ x_{1} = \frac{12 + 22}{10} = \frac{34}{10} = 3.4 \] \[ x_{2} = \frac{12 - 22}{10} = \frac{-10}{10} = -1 \] Now we can find the corresponding \( y \) values by substituting these \( x \) values back into the line's equation: For \( x_{1} = 3.4 \): \[ y = 5 - 2(3.4) = 5 - 6.8 = -1.8 \] For \( x_{2} = -1 \): \[ y = 5 - 2(-1) = 5 + 2 = 7 \] So the coordinates of points \( A \) and \( B \) are \( A(3.4, -1.8) \) and \( B(-1, 7) \). Now we can find the distance between \( A \) and \( B \) using the distance formula: \[ AB = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \] \[ AB = \sqrt{(-1-3.4)^{2}+(7-(-1.8))^{2}} \] \[ AB = \sqrt{(-4.4)^{2}+(8.8)^{2}} \] \[ AB = \sqrt{19.36+77.44} \] \[ AB = \sqrt{96.8} \] \[ AB \approx 9.8 \] Since the distance between \( A \) and \( B \) is approximately \( 9.8 \), which is not equal to the radius of the circle (which is \( 5 \)), we can conclude that \( AB \) is not the diameter of the circle.