## anonymous one year ago What is the equation of the circle with center (2,-5) that passes through the point (-2,10)? a. (x-2)^2+(y-(-5))^2=25 b. (x-(-2))^2+(y-10)^2=241 c. (x-2)^2+(y-(-5))^2=241 d. (x-(-2))^2+(y-10)^2=25 I'm not just asking for an answer. I don't understand and any form of help would be appreciated :)

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1. geerky42

Do you know about center part?

2. geerky42

Saying we have center $$(h,k)$$, then equation would be $$(x-h)^2+(y-k)^2 = r^2$$ Right?

3. anonymous

I think that I might understand what you're getting at.. are you implying that the answer would be "a. (x-2)^2+(y-(-5))^2=25" or would it be "c. (x-2)^2+(y-(-5))^2=241" ...?

4. geerky42

Yeah, we are either at A or C, we don't know yet because we haven't figure out the radius part.

5. geerky42

$$r$$ in $$(x-h)^2+(y-k)^2 = r^2$$ stands for radius. You can find radius by using distance formula on points $$(2,-5)$$ and $$(-2,10)$$

6. geerky42

Distance formula is $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

7. Loser66

You can use directly the general equation of a circle to get r by replace the point into the formula |dw:1432952523897:dw|

8. geerky42

yeah, that's better way to figure out radius.

9. anonymous

(2,-5)=x1 and y1 and (-2,10)=x2 and y2 (-2(-2))^2+(10-(-5))^2 0=25 25 so, would the answer be "a. (x-2)^2+(y-(-5))^2=25" ?

10. geerky42

$$(x_2-x_1)$$ part is supposed to be -4. $$(-2) - 2 = -4$$, right? and $$y_2-y_1$$ is supposed to be 15, since 10 - (-5) = 10 + 5 = 15

11. anonymous

-4^2+15^2 ?

12. geerky42

$$(-4)^2+(15)^2$$, yeah.

13. geerky42

Do you understand?

14. anonymous

I think so... I still get answer choice a as my answer each time I work this out.. is that correct or am I still totally lost?

15. geerky42

Let's just do @Loser66 's way. Plug $$(-2,10)$$ into $$(x-2)^2+(y+5)^2$$.

16. geerky42

And solve. What do you get?

17. anonymous

-4^2+15^2 16+225 241 oh wow..... I see where I messed up! Thank you both so much for helping me and remaining patient :)

18. geerky42

No problem