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1 year ago

The circle below is centered at the point (-3/4) and has a radius of length 3 what is its equation??

2 answers:

6 0

{x-(-3/4)}^2=3^2
(x+3/4)^2=9
{(4x+3)/4}^2=9
(4x+3)^2/16=9
(4x+3)^2=9*16
16x^2+2*4x*3+3^2=144
16x^2+24x+9=144
16x^2+24x+9-144=0
16x^2+24x-135=0

Juli2301 [7.4K]1 year ago

4 0

Answer:

$(x+3)^2+(y-4)^2=9$

Step-by-step explanation:

We are given a circle which is centered at the point (-3,4) and has a radius of length 3.

We have to find the equation of the circle.

The equation of circle centered at (a,b) and radius r is:

$(x-a)^2+(y-b)^2=r^2$

Here a=-3 , b=4 and r=3

Hence, equation of circle is:

$(x+3)^2+(y-4)^2=3^2$

$(x+3)^2+(y-4)^2=9$

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Given:

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