27.0k views
3 votes
If the circle x square + y square +ax + 2y + 4=0. Passes through (2,2) find the centre of the circle.​

2 Answers

6 votes

factor: x^2+y^2+ax+2y+4=0

First plug in the point (2,2) into the equation to find "a".

2^2+2^2+a(2)+2(2)+4=0

2a=-16

a=-8

So now with the value of a we have:

x^2+y^2-8x+2y+4=0

Let rearrange the terms:

x^2-8x+y^2+2y+4=0

Move the constant over to the right side of the equation.

x^2-8x+y^2+2y=-4

Completing the square is tricky but what needs to be done.

x^2-8x+16+y^2+2y+1=13

And then factor the sum and differences of squares.

(x-4)^2+(y+1)^2=13

Divide by 13.

(x-4)^2/13+(y+1)^2/13=0

Now it is in factored form, so we can get the center of the circle.

The center will be at (4,-1).

User Eamon Moloney
by
8.2k points
5 votes

Not a Conic Section its not written writght

User Stefanvds
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.