Question

Find the standard form of the circle x2 + y2 – 6x + 10y + 24 = 0 by completing the square.

Answer 1

Answer:

$(x-3)^2+ (y+5)^2=10$

Step-by-step explanation:

Given the equation of the circle: $x^2 + y^2 - 6x + 10y + 24 = 0$

We wish to express it in a Standard form:

We begin by re-arranging:

$x^2 - 6x + y^2 + 10y = - 24$

Next, divide the coefficient of x by 2, square it and add it to both sides.

Do the same for y.

$x^2 - 6x +(-3)^2+ y^2 + 10y+5^2 = - 24+(-3)^2+5^2$

Next, we factorize

$(x-3)^2+ (y+5)^2 = - 24+9+25\\(x-3)^2+ (y+5)^2=10$

This is the standard form.