Question

a) Find the co-ordinates of the centre, and the length of the radius, of the circle with the equation: ^2 + ^2 + 6 − 4 = 7b) Calculate the co-ordinates of where the circle crosses both axes

Answer 1

`\begin{gathered} a)\text{ coordinates of center of circle (h, k) = (-3, 2)} \\ \text{Length of radius is }\sqrt[]{20} \\ \\ b)\text{ }(0,\text{ }2+\text{ }\sqrt[]{11}),\text{ (}0,\text{ }2-\sqrt[]{11}),\text{ (1, 0), (-7, 0)} \end{gathered}`Step-by-step explanation:
`\begin{gathered} \text{The given equation of circle:} \\ ^2+^2+6-4=7 \\ \\ \text{The above is the general form of the equation of circle:} \\ ^2+^2+2gx+2f+c\text{ = 0} \end{gathered}`

There are different form a circle can be written:

i) the general form (the one with f, g and c)

ii) the standard form that has the vertex (h, k) and radius r

To get the coordinates of the center of circle and the radius, we need to convert to the standard form of circle:

`\begin{gathered} (x\text{ - }h)^2+(y-^{}k)^2\text{ }=r^2 \\ \text{where (h, k) is the center of circle} \\ r\text{ = radius} \end{gathered}`

To find f, g and c from the given equation:

`\begin{gathered} \text{comparing the given equation and the general equation:} \\ 6x\text{ = 2gx} \\ 6\text{ = 2g} \\ g\text{ = 6/2 = 3} \\ \\ -4y\text{ = 2fy} \\ -4\text{ = 2f} \\ f\text{ = -4/2 = -2} \\ \\ ^2+^2+6-4=7\text{ (rewrite this)} \\ ^2+^2+6-4-7\text{ = 0} \\ ^2+^2+2gx+2f+c\text{ = 0} \\ c\text{ = -7} \end{gathered}`

We need to find the center (h, k):

`\begin{gathered} \text{Center of circle: (-g, -f) = (h, k)} \\ \text{converting f and g to center of the circle:} \\ g\text{ = 3} \\ h\text{ = -g = -3} \\ f\text{ = -2} \\ k\text{ = -f = -(-2) = 2} \\ \\ \text{coordinates of center of circle (h, k) = (-3, 2)} \end{gathered}`

Next, let's find the radius of the circle:

The equation in standard:

b) To find the coordinates where the circle crosses both axes, we will equate the y and x coordinate in the equation of circle to zero respectively:

`\begin{gathered} ^2+^2+6-4=7 \\ \text{when x = 0} \\ (0)^2+y^2\text{ + 6(0) - 4y = 7} \\ 0+y^2\text{ + 6(0) - 4y = 7} \\ y^2\text{ - 4y -7 = 0} \\ u\sin g\text{ quadratic formula:} \\ y\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a\text{ = 1, b = -4, c = -7} \\ y\text{ = }\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-7)}}{2(1)} \\ y\text{ = }\frac{4\pm\sqrt[]{16+28}}{2}\text{ = }\frac{4\pm\sqrt[]{44}}{2} \\ y\text{ = }\frac{4\pm\sqrt[]{4*11}}{2}\text{ = }\frac{4\pm\text{ 2}\sqrt[]{11}}{2} \\ y\text{ = }\frac{2(2\pm\text{ }\sqrt[]{11})}{2}\text{ = }2\pm\text{ }\sqrt[]{11} \\ y\text{ = }2+\text{ }\sqrt[]{11}\text{ or }2-\sqrt[]{11} \\ \\ In\text{ coordinate form: (x, y)} \\ (0,\text{ }2+\text{ }\sqrt[]{11}),\text{ (}0,\text{ }2-\sqrt[]{11}) \end{gathered}`

when y = 0

`\begin{gathered} ^2+^2+6-4=7 \\ ^2+(0)^2+6-4(0)=7 \\ ^2+0+6-0=7 \\ ^2+6=7 \\ ^2+6-7=\text{ 0} \\ u\sin g\text{ quadratic formula:} \\ x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{a = 1, b = 6, c = -7} \\ x\text{ = }\frac{-6\pm\sqrt[]{(6)^2-4(1)(-7)}}{2(1)} \\ x\text{ = }\frac{-6\pm\sqrt[]{36+28}}{2}\text{ = }\frac{-6\pm\sqrt[]{64}}{2} \\ x\text{ = }(-6\pm8)/(2) \\ x\text{ = }(-6+8)/(2)\text{ or }(-6-8)/(2) \\ x\text{ = 2/2 or -14/2} \\ x\text{ = 1 or -7} \\ \\ In\text{ coordinate form: (x, y)} \\ (1,\text{ 0) , (-7, 0)} \end{gathered}`

`\begin{gathered} \text{Hence, the coordinates of where the circle crosses both axes:} \\ (0,\text{ }2+\text{ }\sqrt[]{11}),\text{ (}0,\text{ }2-\sqrt[]{11}),\text{ (1, 0), (-7, 0)} \end{gathered}`