Question

Find the equation to the circle which:

• A. (i) touches the axis of x and passes through the two points (1, -2) and (3, -4).
• B. 4i) touches the axis of y at origin, and passes through the point (b, c).

Answer 1

Answer: (B)

The equation of the circle is $bx^2 + by^2 - (b^2 + c^2)x = 0$.

Answer 2

A circle touching the y-axis at the origin has its center on the x-axis at $(h,0)$ and radius $|h|$. Its equation is $(x-h)^2 + y^2 = h^2$, which simplifies to $x^2 + y^2 - 2hx = 0$. The condition that the circle passes through point $(b, c)$ implies $b^2 + c^2 - 2hb = 0$. If $b \neq 0$, we solve for $h$ as $h = \frac{b^2 + c^2}{2b}$ and substitute it back into the circle's equation to obtain the final result. If $b=0$, then $c$ must be $0$, and the point is the origin, leading to a family of circles. The unique equation is derived assuming $b \neq 0$.