Given:
The center of the circle is at (0,0).
The circle passes through the point A(4,-3).
To find:
The radius and the equation of the circle.
Solution:
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
The radius of the circle is the distance between the points (0,0) and (4,-3).
[tex]r=\sqrt{(4-0)^2+(-3-0)^2}[/tex]
[tex]r=\sqrt{(4)^2+(-3)^2}[/tex]
[tex]r=\sqrt{16+9}[/tex]
[tex]r=\sqrt{25}[/tex]
[tex]r=5[/tex]
So, the radius of the circle is 5 units.
The standard form of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where, (h,k) is center of circle and r is the radius.
Putting h=0, k=0, r=5, we get
[tex](x-0)^2+(y-0)^2=5^2[/tex]
[tex]x^2+y^2=25[/tex]
Therefore, the radius of the circle is 5 units and the equation of the circle is [tex]x^2+y^2=25[/tex].
