Question

The center of a circle with (1,2)and(7,-4)as end points of the diameter is​

Answer 1

(x−h)2+(y−k)2=r2 ( x - h ) 2 + ( y - k ) 2 = r 2 is the equation form for a circle with r radius and (h,k) as the center point. In this case, r=√26 and the center point is (2,7) . The equation for the circle is (x−(2))2+(y−(7))2=(√26)2 ( x - ( 2 ) ) 2 + ( y - ( 7 ) ) 2 = ( 26 ) 2 .

Answer 2

The center of a circle with (1, 2), and (7, -4) as diameter points is (4, -1).

Given,

Endpoints of diameter = (1, 2) and (7, -4)

To Find,

The center of the circle.

Solution,

Let the center of the circle be (x, y).

We know that the center of the circle is the midpoint of the diameter so,

⇒ x = (7 + 1)/2

⇒ x = 4

And,

⇒ y = (-4 + 2)/2

⇒ y = -1

∴ (x, y) = (4, -1)

Hence, the center of the circle with (1, 2), and (7, -4) as diameter points is (4, -1).

#SPJ2