Question

Prove that the point (-1,2) is the centre of a circle passing through the points (-3,1),
(5,9), (8,0) and (6,8). Find the radius of this circle.​

Answer 1

Answer:

Center: (-1,2)

Radius = \sqrt{85}

85

Step-by-step explanation:

A circle passing through the points (-3,11),(5,9),(8,0) and (6,8)

Let center of circle be (h,k)

Equation of circle: (x-h)^2+(y-k)^2=r^2(x−h)

2

+(y−k)

2

=r

2

using passing point to find r and centre

For (-3,11)

(-3-h)^2+(11-k)^2=r^2(−3−h)

2

+(11−k)

2

=r

2

-------------(1)

For (5,9)

(5-h)^2+(9-k)^2=r^2(5−h)

2

+(9−k)

2

=r

2

--------------(2)

For (8,0)

(8-h)^2+(0-k)^2=r^2(8−h)

2

+(0−k)

2

=r

2

--------------(3)

Now solve system of equations (1), (2) and (3) and we get

Subtract eq(1) - eq(2)

(-3-h)^2-(5-h)^2+(11-k)^2-(9-k)^2=r^2-r^2(−3−h)

2

−(5−h)

2

+(11−k)

2

−(9−k)

2

=r

2

−r

2

(2-2h)(-8)+(20-2k)(2)=0(2−2h)(−8)+(20−2k)(2)=0

4h-k=-64h−k=−6 -----------------------------(4)

Subtract eq(3) - eq(2)

(8-h)^2-(5-h)^2+(0-k)^2-(9-k)^2=r^2-r^2(8−h)

2

−(5−h)

2

+(0−k)

2

−(9−k)

2

=r

2

−r

2

(13-2h)(3)+(9-2k)(-9)=0(13−2h)(3)+(9−2k)(−9)=0

h-3k=-7h−3k=−7 -----------------------------(5)

Solve eq(4) and eq(5) for h and k

substitute k from equation 4 to eq(5)

h-3(4h+6)=-7h−3(4h+6)=−7

h=-1h=−1

k=2k=2

Center: (-1,2) hence proved

For center put center and passing point in equation (3)

(h,k) ⇔ (-1,2) , Passing point (6,8)

(6+1)^2+(8-2)^2=r^2(6+1)

2

+(8−2)

2

=r

2

r^2=49+36r

2

=49+36

r=\sqrt{85}r=

85

Hence, The center of given circle is (-1,2) and radius is \sqrt{85}

85

I hope you helpful this answer

Answer 2

Answer:

$r = 25$

$hope \: its \: help \: you$