What-is-the-equation-of-the-circle-passing-though-2-4-and-4-6-its-center-lies-on-the-line-3x-2y-20-0-

Question Number 136781 by bramlexs22 last updated on 25/Mar/21

What is the equation of the circle passing though (-2,-4) and (4,6) its center lies on the line 3x- 2y+20=0

Answered by EDWIN88 last updated on 26/Mar/21

Let P (- 2, - 4) and Q (4, 6), and midpoint

PQ is T (\frac{4 - 2}{2}, \frac{6 - 4}{2}) = T (1, 1) .

let C (a, b) be a center point the circle .

We know that vector PQ ⊥ CT; then

{\begin{cases} PQ = (4 - (- 2), (6 - (- 4)) = (6, 10) \\ CT = (1 - a, 1 - b) \end{cases}

PQ . CT = 0; 6 (1 - a) + 10 (1 - b) = 0

\Rightarrow 16 - 6 a - 10 b = 0; 3 a + 5 b = 8 \dots (i)

Because the center point the circle lies

on the line 3 x - 2 y + 20 = 0, we get

\Rightarrow 3 a - 2 b = - 20 \dots (ii)

Now we can solving for a and b, gives

3 a + 5 b = 8

3 a - 2 b = - 20

7 b = 28 {\begin{cases} b = 4 \\ a = - 4 \end{cases}, now we want to

compute the radius = ∣ CP ∣ = \sqrt{{(- 2 + 4)}^{2} + {(- 4 - 4)}^{2}}

radius = \sqrt{4 + 64} = \sqrt{68 .}

Finally the equation of the circle is equal to

{(x + 4)}^{2} + {(y - 4)}^{2} = 68 or

x^{2} + y^{2} + 8 x - 8 y - 34 = 0

Commented by EDWIN88 last updated on 26/Mar/21

Answered by mr W last updated on 26/Mar/21

s a y c e n t e r o f c i r c l e i s (u, v), r a d i u s r

3 u - 2 v + 20 = 0 \dots (i)

{(u + 2)}^{2} + {(v + 4)}^{2} = r^{2} \dots (i i)

{(u - 4)}^{2} + {(v - 6)}^{2} = r^{2} \dots (i i i)

(i i) - (i i i) :

3 u + 5 v - 8 = 0 \dots (i v)

(i v) - (i) :

7 v - 28 = 0 \Rightarrow v = 4

u = \frac{8 - 5 v}{3} = \frac{8 - 20}{3} = - 4

r^{2} = {(- 4 + 2)}^{2} + {(4 + 4)}^{2} = 68

e q n . o f c i r c l e :

{(x + 4)}^{2} + {(y - 4)}^{2} = 68

Commented by otchereabdullai@gmail.com last updated on 27/Mar/21

nice!