Find the equation of the circle which passes through the origin and cuts off intercepts $3$ and $4$ from the positive parts of the axes respectively.
My Attempt,
The circle cuts the axis of $X$ at $(3,0)$ and the axis of $Y$ at $(0,4)$. Let $r$ be the radius of the circle. Since the circle passes through the origin, $$r=\sqrt {h^2+k^2}$$,
Where, $(h,k)$ are the co ordinates of the centre of the circle.
On
Do it geometrically.
Bisect the line segment from $(0,0)$ to $(3,0)$. And the line segment from $(0,0)$ to $(0,4)$.
The center of the circle is obviously $(1.5,2)$.
The radius of the circle is $\sqrt {1.5^2 + 2^2}$.
The equation is thus: $$(x-1.5)^2+(y-2)^2=1.5^2 + 2^2$$
Simplify and expand to get the desired form.
Let the equation of the circle be $$x^2+y^2+2gx+2fy+c=0$$
Now it passes through $(0,0);(3,0);(0,4)$
We have three unknowns with there conditions, right?