Find the centre of a circle that passes through the points D (-5,4), E (-3,8), and F (1,6).

Question

I really need help.

 

Answer 1

Gven points are D(-5,4), E(-3,8), F(1,6)

Standerd form of a crcle is (x-h)^2+(y-k)^2 = r^2 -------> (1)

Where (h,k)is center of circle.

Now substitute the each point in (1).

(-5-h)^2+(4-k)^2 = r^2 -------> (2)

(-3-h)^2+(8-k)^2 = r^2 ---------> (3)

(1-h)^2+(6-k)^2 = r^2 ----------> (4)

From (2) and (3) we can able to write (-5-h)^2+(4-k)^2 = (-3-h)^2+(8-k)^2

(-5-h)^2-(-3-h)^2 = (8-k)^2-(4-k)^2

Here we can use the formula  a^2-b^2 = (a-b)(a+b)

[(-5-h)+(-3-h)][(-5-h)-(-3-h)] = [(8-k)+(4-k)][(8-k)-(4-k)]

[-5-h-3-h][-5-h+3+h] = [8-k+4-k][8-k+4+k]

(-8-2h)(-2) = (-2k+12)(12)

16+4h = -24k+144

4h+24k = 144-16

4h+24k = 128

2h+12k = 64 

Multple to each side by 3

6h+36k = 192 -----------> (5)

From (2) and (4) we can able to write (-5-h)^2+(4-k)^2 = (1-h)^2+(6-k)^2

(-5-h)^2-(1-h)^2 = (6-k)^2-(4-k)^2

Here we can use the formula  a^2-b^2 = (a-b)(a+b)

[(-5-h)-(1-h)][(-5-h)+(-1-h)] = [(6-k)-(4-k)][(6-k)+(4-k)]

[-5-h-1+h][-5-h-1-h] = [6-k-4+k][6-k+4-k]

(-6)(-2h-6) = (2)(-2k+10)

12h+36 = -4k+20

Add 4k to each side.

12h+36+4k = -4k+4k+20

Subtract 36 from each side.

12h+4k+36-36 = 20-36

12h+4k = -16

3h+k = -4

Multiple to each sde by negitive 2

-6h-2k = 8 ---------->(6)

Now we add the equations (5)&(6).

6h+36k = 192

-6h-2k = 8

__________

34k = 200

k = 200/34

k = 100/17

3h+100/17 = -4

3h = -4-100/17

3h = -168/17

h = -168/51

h = -56/17

Center (h,k) = (-56/17,100/17)

Comments

The solution is wrong,

the center of the circle C(h, k) passing through the given points is (- 2, 5).

Answer 2

The points are D(- 5, 4), E(- 3, 8), and F(1, 6).

Let the center and radius of the circle be C(h, k) and r.

|DC| = |EC| = |FC|         (Since, the distance from the center and given three points are equal(radius))

|DC|² = |EC|² = |FC|²

(h + 5)² + (k - 4)² = (h + 3)² + (k - 8)² = (h - 1)² + (k - 6)²

(h + 5)² + (k - 4)² = (h + 3)² + (k - 8)² → h + 2k = 8  -------> (1)

(h + 3)² + (k - 8)² = (h - 1)² + (k - 6)² → 3h + k = - 1 -------> (2)

Solve the eqn' s (1) & (2), for (h, k).

Multiply eq (2) by 2, and write the equations in column form, then subtract them, and solve for h.

h + 2k = 8

6h + 2k = - 2

( - )___________

- 5h = 10

⇒ h = 10/- 5 = - 2.

Substitute h = - 2 in eq (1), and solve for k.

- 2 + 2k = 8

2k = 8 + 2 = 10

⇒ k = 10/2 = 5.

Therefore, the center of the circle C(h, k) passing through the given points is (- 2, 5).