What is the standard form of the equation of a circle that has its center at (-2, -3) and passes through the point (-2, 0)?
(A) (x + 2)^2 + (y + 3)^2 = 9 (B) (x − 2)^2 + (y − 3)^2 = 16 (C) (x − 2)^2 + (y − 3)^2 = 4 (D) (x + 2)^2 + (y + 3)^2 = 16 (E) (x − 2)^2 + (y + 3)^2 = 9
formula for a circle with radius \(r\) and center at \((h,k)\) is \[(x-h)^2 + (y-k)^2 = r^2\]You don't know \(r\) yet, but you do know the center. What is the first pass at the formula going to be?
(x-(-2)^2 + (y-(-3)^2=r^2
Yes. Can you simplify that any more? by the way, you need some more parentheses — \((x-(-2))^2 + (y-(-3))^2 = r^2\) Next step is to find the value of \(r\). There are two ways that come to mind how you could do this. 1) find the length of the radius from \((-2,-3)\) to \((-2,0)\) with the distance formula \[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]which gives the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) 2) just plug in your known point \((-2,0)\) as \((x,y)\) in the formula and find the value of \(r\)
Personally, I think the second one is easier. You'll actually end up with the value of \(r^2\), and that's what you want to use in the formula, so no need to solve it for \(r\)
(x+2)^2 + (y+3)^2=r^2 Does it simplify to this?
Yes, good job not getting tripped up by the - signs in the formula.
Ok now do i raise (x+2),(y+3) by the power of 2?
Thank you @whpalmer4 turns out i didn't have to do this problem. Although i have one pretty similar, would you mind?
(x + 5.5)2 + (y + 4)2 = 3.5 , (x − 5.5)2 + (y − 4)2 = 3.5 I know the answer is between these two because the center is at (5.5,4), and the radius is 3.5.
Don't get tripped by the negative sign in the formula!
Do you recall the formula?
For the equation of a line?
Of a circle.
Duh, sorry haha.
(x-a)2 + (y-b)2 = r2 ?
Yep, what's a, b, and r?
a=5.5, b=4, c=1.75 ?
wait no.
Why 1.75?
3.5* I divided by 2 for some reason.
Now plug in the values?
(x-5.5)^2+(y-4)^2=3.5^2
Thank you @kc_kennylau
I appreciate it!!
Join our real-time social learning platform and learn together with your friends!