WILL GIVE MEDAL!! What is the general form of the equation for the given circle? x2+y2-8x-8y+23=0 x2+y2-8x-8y+32=0 x2+y2-4x-4y+23=0 x2+y2+4x+4y+9=0
I think you'd actually save time if you took a second to understand the method in one of the solutions you've already asked for, rather than expecting people to do every question for you. You also might actually learn something.
im with you Newton 😂 but there is no harm as long as there is medals😂
Here a=radius of circle (h,k)=coordinates of centre of circle
(x-h)^2+(y-k)^2=a^2
Please help me
Can u tell me the distance AB=?
How do I do that?
there is something missing i thing i spent thinking 5min on it :s recheck ur ques
think*
(AB)^2=(x1-x2)^2+(y1-y2)^2
substitute the given points in each equation and check
If OAB is a right angled triangle. Then we will b able to find out radius of given circle
Radius a=?
no we cannot there are 2 unknowns sir
\[x^2+y^2-8x-8y+23=0\] 1. put x=4,y=7 2. put x=7,y=4 and see if it satisfies the given eq.
im telling you there is something missing
@surjithayer your just plugging in lameee we want the real math
if it does not satisfy then try second eq. and so on.
a^2+a^2=(AB)^2
a=?
|dw:1423505610221:dw|
I think we can solve it
We hv 2 equations nd we hv to fond out 2 unknown h,k
Just follow my procedure
Sorry i lying on my bed for sleep. Its midnight here
what about radius? so three unknowns h,k,r and only two equations .we can prove h=k but what about r?
you are assuming OA perpendicular OB.
Ya i guess OAB is a right angled triangle
but it is not given it is a right angled triangle.
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