Question

Find the centre and radius of -6x+x^2=97+10y-y^2

Answer 1

Centre = (3, 5)

Radius =
`√(131)`

Explanation:

Given equation of a circle:

`-6x + x^2 = 97 + 10y - y^2`

To find the centre and radius of the given equation of a circle, rewrite it in standard form.

`\boxed{\begin{minipage}{6.3cm}\underline{Equation of a Circle - Standard Form}\\\\$(x-h)^2+(y-k)^2=r^2$\\\\where:\\ \phantom{ww}$\bullet$ $(h, k)$ is the centre. \\ \phantom{ww}$\bullet$ $r$ is the radius.\\\end{minipage}}`

First, rearrange the equation so that the terms in x and y are on the left side and the constant is on the right side:

`x^2 - 6x + y^2 - 10y = 97`

Complete the square for the x and y terms by adding the square of half the coefficient of the term in x and y to both sides:

`\implies x^2 - 6x +\left((-6)/(2)\right)^2+ y^2 - 10y +\left((-10)/(2)\right)^2= 97+\left((-6)/(2)\right)^2+\left((-10)/(2)\right)^2`

Simplify:

`\implies x^2 - 6x +\left(-3\right)^2+ y^2 - 10y +\left(-5\right)^2= 97+\left(-3\right)^2+\left(-5\right)^2`

`\implies x^2 - 6x +9+ y^2 - 10y +25= 97+9+25`

`\implies x^2 - 6x +9+ y^2 - 10y +25= 131`

Now we have created two perfect square trinomials on the left side of the equation:

`\implies (x^2 - 6x +9)+ (y^2 - 10y +25)= 131`

Factor the perfect square trinomials:

`\implies (x-3)^2+ (y-5)^2= 131`

If we compare this equation with the standard form, we see that the centre of the circle is (3, 5) and its radius is the square root of 131.

Therefore:

• centre = (3, 5)
• radius =
  `√(131)`