What is the equation of a circle with its center at (−6,−3) and a radius of 12?
2 answers:
Answer:
x² + y² + 12x +6y - 99 = 0
Step-by-step explanation:
<u>G</u><u>iven </u><u>:</u>
- Centre = (-6,-3)
- Radius = 12 ,
<u>Using</u><u> </u><u>the </u><u>Standard</u><u> equation</u><u> of</u><u> </u><u>circle</u><u> </u><u>,</u><u> </u>
- ( x - h)² + (y-k)² = r²
- ( x +6)² + (y+3)² = 12²
- x² + 36 + 12x + y² + 9 + 6y = 144
- x² + y² + 12x + 6y +45-144 = 0
- x² + y² + 12x +6y - 99 = 0
Answer:
(x+6)^2 + ( y +3)^2 = 144
Step-by-step explanation:
The equation of a circle is given by
(x-h)^2 +(y-k)^2 = r^2 where (h,k) is the center and r is the radius
(x- -6)^2 + ( y - -3)^2 = 12^2
(x+6)^2 + ( y +3)^2 = 144
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Answer: 5
Step-by-step explanation:
Here is how to solve it with long division and synthetic division.
I’m not sure but I got 8 ^-5
-2(x-6)+7=35
-2x+6+7=35 (Distributive property)
-2x+13=35 (Simplify, combine like terms)
-2x+13-13=35-13
-2x=22
-2x/-2=22/-2
X=-11