Question

Find the equation to the circle passing through the points (0,a), (b,h), and having its centre on x-axis.

Answer 1

$x^2 + y^2 - \left(\frac{b^2 + h^2 - a^2}{b}\right) x - a^2 = 0$

Answer 2

Let the equation of the circle be $(x-h_c)^2 + (y-k_c)^2 = r^2$. Given that the center lies on the x-axis, $k_c = 0$. The equation becomes $(x-h_c)^2 + y^2 = r^2$. The circle passes through $(0,a)$: $(0-h_c)^2 + a^2 = r^2 \implies h_c^2 + a^2 = r^2$. (1) The circle passes through $(b,h)$: $(b-h_c)^2 + h^2 = r^2$. (2) Equating (1) and (2): $h_c^2 + a^2 = (b-h_c)^2 + h^2$. Solving for $h_c$: $h_c^2 + a^2 = b^2 - 2bh_c + h_c^2 + h^2 \implies a^2 = b^2 - 2bh_c + h^2 \implies 2bh_c = b^2 + h^2 - a^2 \implies h_c = \frac{b^2 + h^2 - a^2}{2b}$. (Assuming $b \neq 0$ for a unique circle). Substitute $r^2$ from (1) and $h_c$ into the circle equation $(x-h_c)^2 + y^2 = r^2$: $x^2 - 2xh_c + h_c^2 + y^2 = h_c^2 + a^2$ $x^2 + y^2 - 2xh_c = a^2$. Substitute the value of $h_c$: $x^2 + y^2 - 2x \left(\frac{b^2 + h^2 - a^2}{2b}\right) = a^2$. Simplify to get the final equation: $x^2 + y^2 - x \left(\frac{b^2 + h^2 - a^2}{b}\right) - a^2 = 0$.