Question

A circle has end points (-2,1) and (8,3).

Answer 1

Answer:

A) find its center  = (3, 2)

B) find the radius of the circle = √104

C) find the equation of the circle = x² + y² - 6x - 4y -91 = 0

Step-by-step explanation:

A)- The center must be the mid-points of (-2, 1) and (8, 3).

So, using the equation of mid-point,

$h=\frac{x_{1}+x_{2}}{2} and k=\frac{y_{1}+y_{2}}{2}$

Here, (x₁, y₁) = (-2, 1) and (x₂, y₂) = (8, 3)

Putting these value in above equation. We get,

h = 3 and k = 2

Thus, Center = (h, k) = (3, 2)

B)- For finding the radius we have to find the distance between center and any of the end point.

Thus using Distance Formula,

$Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$Radius =\sqrt{(8+2)^{2}+(3-1)^{2}}$

⇒ Radius = √104 = 2√26

C)- The equation of circle is determined by formula:

$(x-h)^{2}+(y - k)^{2} = r^{2}$

where (h, k ) is center of circle and

r is the radius of circle.

⇒ (x - 3)² + (y - 2)² = 104

⇒ x² + y² - 6x - 4y -91 = 0

which is the required equation of the circle.