Centre at $(0, 0)$ and which passes through the points $(3, 2)$ and $(1, 6)$ .

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Solution

Let equation of circle whose centre is at $(0, 0)$ be
$(x - 0)^{2} + (y - 0)^{2} = r^{2}$ ['r' being its radius]
or, $x^{2} + y^{2} = r^{2}$
Given the circle passes through $(3, 2)$ then we have,
$3^{2} + 2^{2} = r^{2}$
or $r^{2} = 13$
So equation of the circle is $x^{2} + y^{2} = 13$ .

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Centre at (0,0) and which passes through the points (3,2) and (1,6).

Let equation of circle whose centre is at (0,0) be(x−0)2+(y−0)2=r2 ['r' being its radius]or, x2+y2=r2Given the circle passes through (3,2) then we have,32+22=r2or r2=13So equation of the circle is x2+y2=13.

Centre at $(0, 0)$ and which passes through the points $(3, 2)$ and $(1, 6)$ .

Let equation of circle whose centre is at $(0, 0)$ be
$(x - 0)^{2} + (y - 0)^{2} = r^{2}$ ['r' being its radius]
or, $x^{2} + y^{2} = r^{2}$
Given the circle passes through $(3, 2)$ then we have,
$3^{2} + 2^{2} = r^{2}$
or $r^{2} = 13$
So equation of the circle is $x^{2} + y^{2} = 13$ .