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Write the standard form of the equation of the circle that passes through (2,3) and has a center at (0,0)
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equation of the circle with centre at (0,0) is given by x^2 + y^2 =r^2 where r is the radius
since the centre of the required circle is known to be (0,0) and that it is also given that the circle passes through (2,3), so the distance between the points (0,0) and (2,3) gives the radius of the circle...
\[\huge (x-a)^2+(y-b)^2=r^2\] Where (a,b) is the center.
i hope you can proceed from here...
so it (x-2)^2+(y-3)^2=r^2
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oops I meant *so is it
not quite since to circle has the origin at the centre its \[x^2 + y^2 = r^2\] to find the value of r^2 substitute x = 2 and y = 3 then evalaute
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