Question

find the equation of a circle whose Centre is(- 3,1 )and which passes through the point (5, 2) ​

Answer 1

x + 8y - 5 = 0

Equation of the line , when two points are given is

$y - 1 = \frac{2 - 1}{ - 3 - 5} (x + 3) \\ - 8y + 8 = x + 3 \\ x + 8y - 5 = 0$

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Answer 2

Answer:

equation of circle is: x² + y² + 6x - 2y - 55 = 0

Step-by-step explanation:

given that:

center of a circle is ( -3 , 1 )

and the circle passes through the point (5 , 2)

so the distance between the points (-3 , 1) and (5 , 2) is radius 'r'.

according to formula of circle:

if (h , k) is center , the circle equation is :

(x - h)² + (y - k)² = r²

r = distance between points (-3 , 1) and (5 , 2)

using distance formula ;

$r = \sqrt{(-3-5)^{2}+(1-2)^{2} }$

$r^2 = (-8)^2+(-1)^2$

$r^2=64+1=65$

also center (h , k) = ( -3 , 1)

equation of circle is:

(x - (-3))² + (y - 1)² = 65

(x + 3)² + (y - 1)² = 65

x² + 9 + 6x + y² + 1 - 2y = 65

x² + y² + 6x - 2y - 55 = 0