Please help me out its urgent, “find the equation of the circle that passes through the points (0,0) (2,0)&(3,-3) About the author Bella
Answer: Three points on the circle are:−A≡(0,2),B≡(3,0),C≡(3,2) Let P≡(x,y) be its centre, then ∣PA∣=∣PB∣=∣PC∣ ⇒ (x−0) 2 +(y−2) 2 = (x−3) 2 +(y−0) 2 = (x−3) 2 +(y−2) 2 ⇒ x 2 +y 2 +4−4y = x 2 +9−6x+y 2 = x 2 +9−6x+y 2 +4−4y ⇒4−4y=9−6x=9−6x+4−4y Taking first and third parts of the above equation, 4−4y=9+4−6x−4y⇒6x=9⇒x= 6 9 = 2 3 Taking second and third parts of the above equation, 9−6x=9+4−6x−4y⇒4y=4⇒y=1∴P≡( 2 3 ,1) So, radius r=PA= ( 2 3 ) 2 +(1−2) 2 = 4 9 +1 = 2 5 Thus, equation of the circle ;_(x− 2 3 ) 2 +(y−1) 2 =( 2 5 ) 2 ⇒x 2 +y 2 −3x−2y+1− 4 5 + 4 9 =0⇒x 2 +y 2 −3x−2y=0 Step-by-step explanation: Hope it is correct Reply
Answer:
Three points on the circle are:−A≡(0,2),B≡(3,0),C≡(3,2)
Let P≡(x,y) be its centre, then ∣PA∣=∣PB∣=∣PC∣
⇒
(x−0)
2
+(y−2)
2
=
(x−3)
2
+(y−0)
2
=
(x−3)
2
+(y−2)
2
⇒
x
2
+y
2
+4−4y
=
x
2
+9−6x+y
2
=
x
2
+9−6x+y
2
+4−4y
⇒4−4y=9−6x=9−6x+4−4y
Taking first and third parts of the above equation,
4−4y=9+4−6x−4y⇒6x=9⇒x=
6
9
=
2
3
Taking second and third parts of the above equation,
9−6x=9+4−6x−4y⇒4y=4⇒y=1∴P≡(
2
3
,1)
So, radius r=PA=
(
2
3
)
2
+(1−2)
2
=
4
9
+1
=
2
5
Thus, equation of the circle ;_(x−
2
3
)
2
+(y−1)
2
=(
2
5
)
2
⇒x
2
+y
2
−3x−2y+1−
4
5
+
4
9
=0⇒x
2
+y
2
−3x−2y=0
Step-by-step explanation:
Hope it is correct