general form of the equation with center (-1,2) and solution point (0,0)

do i need to find the distance between those two points to find the radius?

Circle i am sure. But when you say solution point you mean that (0,0) is a point on the circle correct?

i dont know. it's just what my book says

and yes it is a circle. that i know

Okay well, i can work with this. The general form of a circle is this: (x-h)² + (y-k)² = r² What you do is plug in the center which in our case is (-1,2) and we get: (x+1)² + (y-2)² = r² also, we have the point (0,0) which is a point on the circle, we plug that into the standard equation of a circle and get: (0+1)² + (0-2)² = r² Now all you have to do is evluate this : (x+1)² + (y-2)² =(0+1)² + (0-2)²

but could i use the distance formula to find the radius? then do everything else? would that be harder?

you see the right hand side of this final equaltion i gave you: (x+1)² + (y-2)² =(0+1)² + (0-2)² once you evluate that, then you have your radius

yeah it's 5

should all that be squared again on the right side? because radius is the equation is r^2

yes to get it in standard form you can square the 5, but the radius is 5

\[\sqrt{(x _{2}-x _{1})^2+(y _{2}-y _{1})^2}\]

to use the distance formula simply plug in your given points and solve, in this case (x1,x2)=(-1,0) and (y1,y2)=(2,0)

sorry i was lookin at my work. i just haven't got that far. i do good answers when i understand and figure out the problem.

you dont understand yet? how can i make it clearer for you?

well i got it, but, my raidus was \[\sqrt{5}\] and with the rest of my work it was right... so i guess i might have missed something. idk

now the radius is 5

the radius squared is 5

right isnt that what you wanted

\[r^2=1^2+2^2=5\]as in pythagoras.

not really. thats why i was so confused

then write \[(x+1)^2+(y-2)^2=5\]

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