Find the center and radius of the following circles x ^ 2 + y ^ 2 - 4x - 8y - 41 = 0, 3x ^ 2 + 3y ^ 2 + 6x - 12y - 1 = 0
Answers
Answered by
1
Answer:
I don't know the answer sorry
Answered by
3
Step-by-step explanation:
Answer
Open in answr app
Open_in_app
3x
2
+3y
2
−6x+4y−1=0⇒[(
3
x)
2
−2×
3
×
3
×x+(
3
)
2
]+[(
3
y)
2
+2×
3
×
3
2
y+(
3
2
)
2
]−3−
3
4
−1=0
⇒(
3
)
2
+(
3
y+
3
2
)
2
−
3
16
=0
3(x−1)
2
+3(y+
3
2
)
2
−
3
16
=0
(x−1)
2
+(y+
3
2
)
2
=
3×3
16
=(
3
4
)
2
center (1,
3
−2
) radius =
3
4
Similar questions