x^2+y^2-10x+20y-44=0

Question

reduce to standard form,find the coordinates of the center, vertex or vertices,foc,latus ******, direction of opening.

Answer 1

The equation x ^2 + y ^2 - 10x + 20y - 44 = 0

To change the expression into a perfect square trinomials add (half the x coefficient)² and

(half the y coefficient)² to each side of the expression.

x  coefficient = - 10 then (half the x coefficient)² = 25

y  coefficient = 20 then (half the y  coefficient)² = 100

So add 100 , 25 to each side.

x ^2 + y ^2 -10x + 20y - 44 + 100 + 25= 100 + 25

x ^2 - 10x + 25 + y ^2 + 20y + 100 = 125 + 44

(x - 5)^2 + (y + 10)^2 = 169

(x - 5)^2 + (y - (-10))^2 = (13)^2

Compare it to standard form of circle ( x - h )^2+( y - k )^2 = r ^2

Coordinates of center is (h , k ) = (5, - 10)

Foci is also center of circle = (5, - 10)

r  = 13

In a circle latus rectum is diameter (2r ) = 26

A circle has infinite number of vertices because it contains a continuously rounded perimeter.