Equation of a Circle problems?

Question

Find the equations of the circles:
a) with the centre (-2,3) and radius 6
b) with centre (3,0) and which passes through the point (5,4)
c) with centre (3,-1) and which touches the y-axis.
d) find the centre and radius of the circle x^2 + y^2 - 4x + 8y - 80

Explanations would be very helpful because I mainly want to understand how it is done.

Answer 1

a) with the cent re (-2,3) and radius 6
Ans:
(x+2)^2 + (y-3)^2=6^2
simplify
x^2 +y^2 +4x -6y - 23=0..................i
b) with cent re (3,0) and which passes through the point (5,4)
Ans:
(x-3)^2 + (y-0)^2=r^2
x^2 +y^2 -6x +9 =r^2
Since point(5,4) passes through it
25 +16-30 +9 =r^2
20=r^2
Therefore equation of circle is
x^2 +y^2 -6x +9 =20
x^2 +y^2 -6x -11 =0................ii

c) with cent re (3,-1) and which touches the y-axis.
Ans:
Since it touches y-aix then radius =x -coordinate of cent re=3^2=9
(x-3)^2 + (y+1)^2=3^2
x^2 +y^2 -6x +2y +1=0.........................iii

d) find the cent re and radius of the circle x^2 + y^2 - 4x + 8y - 80
Ans;
x^2 + y^2 - 4x + 8y - 80=0
(x-2)^2 + (y+4)^2=10^2
cent re are(2,-4) radius =10..........iv

Answer 2

The equation of a circle centered at the origin is: x^2 + y^2 = r^2, where r is the radius. To move the circle around, you just adjust your x and y.

For your first problem, the equation will be (x+2)^2 + (y-3)^2 = 6^2.
To see why this works, play with the equation a bit. Obviously, the circle in question has to include the points (-8,3), (-2, 9), (4, 3) and (-2,-3). Plug those coordinates into the equation and see what happens.

For the second problem, the equation will be (x-3)^2 + (y-0)^2 = r^2, where r is the distance between (3,0) and (5,4). To get that distance, just use Pythagoras: the radius is the hypotenuse of a triangle with leg lengths (5-3) and (4-0).

For the third problem, the radius obviously has to be 3. At the point where the circle touches the y-axis, the x-coordinate has to be zero. If you understand the first two problems, you should be able to handle this one once you know the radius.

For the fourth problem... hmm, did you leave out an "equals zero" at the end of the equation? Naughty, naughty - you can't do that! The equals sign is arguably the most important part of any equation! Anyway, you need to complete some squares. To get a square out of x^2 - 4x, you need a +4 term at the end, since (x^2 - 4x +4) = (x-2)^2. Similarly, (y^2 + 8y + 16) = (y+4)^2.
Therefore, rearrange your equation as (x^2-4x+4) + (y^2 + 8y +16) - 100 = 0
That turns into (x-2)^2 + (y+4)^2 -100 = 0
So (x-2)^2 + (y+4)^2 = 100.
You can figure out the center and radius from there :-)

Answer 3

a)(x+2)^2+(y-3)^2=36

b)(x-3)^2+y^2=r^2
passes through (5,4)
so (5-3)^2+4^2=r^2
r^2=18
so the equation
(x-3)^2+y^2=18

c)(x-3)^2+(y+1)^2=(-1)^2

d)centre=-(-4/2) and -(8/2)
=(2,-4)
radius rt(4+16+80)=10