# Find the equation of the circle that passes through (8, 0), (0, 6) and (0, 0).

*print*Print*list*Cite

We start with the general equation of the circle: x^2 + y^2 + ax + by + c = 0

Now substitute the values for the coordinates from the points the circle passes through

For (8, 0)

we have 64 + 0 + 8x + 0 + c = 0 … (1)

For (0, 6)

we have 0 + 36 + 0 + 6y +c = 0… (2)

For (0, 0)

we have 0 + 0 + 0 +0 +c = 0… (3)

So, from (3) we get c = 0

substituting this in (2)

=> 36 + 6y = 0

=> y = -36/6

=> y = -6

substituting c= 0 in (1)

64 + 0 + 8x + 0 + c = 0

=> 64 + 8x = 0

=> x = -64 / 8

=> x = -8

**Therefore the required equation of the circle is x^2 + y^2 – 8x – 6y = 0**

Let the equation of the circle be (x-h)^2+9y-k)^2 = r^2.

Since this passes through the given 3 points , we have:

(0,0):

(0=h)^2+(0-k)^2 = r^2 . Or h^2 +k^2 = r^2.

(8,0):

(8-h)^2 +((0-k)^2 = h^2 +k^2 .

64-16h+ h^2+k^2 = h^2+k^2 . h^2+k^2 on both sides get cancelled.

=> 64 = 16h. So h = 4.

(0,6):

(0-h)^2+(6-k)^2 = h^2+k^2 .

h^2+36-12k+k^2 = h^2+k^2, h^2+k^2 on both sides get cancelled.

=> 36 = 12k. So k = 3.

= r^2 = h^2+k^2 = 4^2+3^2 = 25.

Therefore the equation of the circle is (x-4)^2+(y-3)^2 = 25.

x^2-8x+4^2 +y^2-12y+9=25

x^2+y^2-8x-6y = 0 is the required circle which passes through (8, 0), (0, 6) and (0, 0).

We'll use the determinant:

x^2 + y^2 x y 1

det A = 8^2 + 0^2 8 0 1 = 0

0^2 + 6^2 0 6 1

0^2 + 0^2 0 0 1

x^2 + y^2 x y 1

det A = 64 8 0 1 = 0

36 0 6 1

0 0 0 1

We'll compute the determinant considering the last row, since it has a lot of zeroes.

det A = (-1)^(4 + 4)*1*det A1

x^2 + y^2 x y

det A1 = 64 8 0

36 0 6

det A1 = 48(x^2 + y^2) - 288y - 384x

**The equation of the circle that passes through the given points is:**

**48x^2 + 48y^2 - 288y - 384x = 0**