Given,
The center of the circle is (0,0).
The point on the circle is (6,8).
Required:
The radius of the circle and standard equation of the circle.
a)The radius of the circle is calculated as,
[tex]\begin{gathered} Radius\text{ =}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt{(6-0)^2+(8-0)^2} \\ =\sqrt{6^2+8^2} \\ =\sqrt{36+64} \\ =\sqrt{100} \\ =10 \end{gathered}[/tex]The radius of the circle is 10.
b)The standard equation of the circle is,
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (x-0)^2+(y-0)^2=10^2 \\ x^2+y^2=100 \end{gathered}[/tex]Hence, the equation of the circle is x^2+y^2=100.
c) The point on the third quadrant lies on the circle.
In the third quadrant both x and y values are negative.
So, the point lies in third quadrant is (-6, -8).
Hence, point on third quadrant is (-6, -8).