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.
Answers
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
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Answer: