Equation of Circle?

Question

find the equation of the circle which passes through the point A(0,1) , B(3,-2) and has its centre lying on the line y=x-2

please present workings

There is no information missing from this question. The final answer is its x^2+y^2-2x+2y-3=0 .

Answer 1

The circle has its center at (1.5, -.5). The radius is
sqrt(1.5^2+1.5^2), so r^2 = 4.5.

Hence equation is (x-1.5)^2 +(y+.5)^2 = 4.5

Answer 2

General equation of a circle, (x-a)^2 + (y-b)^2 = r^2
where (a,b) is the centre of the circle.
Because the centre lie on y=x-2, then the coordinates of the centre are (a, a-2). We get :

(x-a)^2 + (y-(a-2))^2 = r^2
(x-a)^2 + (y-a+2)^2 = r^2
Well, the next step seems to be impossible. Because so many variable here. Maybe you forgot to mention the radius of the circle?

Answer 3

let C(h,k) be the center of the circle...
then C(h,k) is equidistant to points A and B. .that is, AC=BC
the distance between two points (x1,y1) and (x2,y2) is
d = [(x1-x2)^2 + (y1-y2)^2]^(1/2)

since AC=BC,
[(0-h)^2 + (1-k)^2]^(1/2) = [(3-h)^2 + (-2-k)^2]^(1/2)
squaring both sides of the equation to eliminate the radical sign. we have
[(0-h)^2 + (1-k)^2] = [(3-h)^2 + (-2-k)^2]
expressing the above as sum of terms,
h^2 + 1 - 2k + k^2 = 9 - 6h + h^2 + 4 + 4k + k^2
combining like terms,
6h - 6k - 12 = 0
dividing the equation by 6 to reduce it,
h - k - 2 = 0 let this be equation (1)...this is the equation of the line containing all the points equidistant to A and B.
this line passes through the center of the circle. therefore, if the center is also contained at the line y = x - 2, we may get the intersection of this line and equation (1). they both contain the center as a point...
using (h,k) as points of the line y = x - 2, we may write it as
k = h - 2 or h - k - 2 = 0...

uh oh...are you sure there is no other line given? coz my equation (1) is exactly the same line as your given line. i find the given insufficient...

Answer 4

Centre: (x, x-2)

x² + (x-2-1)² = (x-3)² + (x-2+2)²
x² + (x-3)² = (x-3)² + x²

x can have any value

There are an infinite number of circles satisfying the required conditions.

Let x=0
y = -2

r² = (0 + (2+1)²) = 9

x² + (y+2)² = 9
x² + y² + 4y + 4 = 9
x² + y² + 4y - 5 = 0 is one such circle

Answer 5

table certain factor is whilst dy/dx=0. So dy/dx = 3x^2 +2ax -15 = 0. If x=-a million on the table certain factor, dy/dx = 2-2a-15=0. So a=-13/2. putting returned into y, y= -a million -13/2*(a million) -15(-a million) + b = 12. paintings out b from that. Then do dy/dx returned to paintings out the recent table certain factor. you will optimistically get 2 table certain factors, one in each and every of them being (-a million,12)!

Answer 6

Here it's impossible because the line : y=x-2
is the mediatrice of [AB]

so the number of solutions is infinit