Question

Find an equation of the circle with center at the origin and passing through (3,-5) in the form of (x-a)^2+(y-b)^2=c

Answer 1

General Idea:

`Equation \; of \; the \; circle \; is \; given \; by:\\(x-h)^2+(y-k)^2=r^2\\Where:\\(h,k) \; is \; the \; centre\\r \; is \; the \; radius`

Applying the concept :

`Given \; Center \; of \; the \; circle \; at \; the \; origin,\\So \; (h, k)=(0,0)\\\\Circle \; passing \; through \; (3,-5)\\So \; Radius \; is \; the \; distance \; between \; (0, 0) \& (3,-5)`

`R=√((x_2-x_1)^2+(y_2-y_1)^2)\\\\Substituting \; (0,0) \; \& \; (3,-5)\\\\R=√((-5-0)^2+(3-0)^2)=√(25+9)=√(34)`

Conclusion:

By substituting the above information the equation of circle, we get...

`(x-0)^2+(y-0)^2=(√(34))^2\\(x-0)^2+(y-0)^2=34`