圆心在 $x$-axis 上，圆又 touches the $y$-axis，所以圆一定与 $y$-axis 相切。因为它还要经过 $(5,6)$，圆应在 $x$-axis 上下对称，并且只在右侧接触 $y$-axis。

Sketch a circle which:

• touches the $y$-axis at the origin,
• is symmetrical about the $x$-axis,
• lies in quadrants 1 and 4,
• passes through the point $(5,6)$.

设圆心为 $(a,0)$。因为圆 touches the $y$-axis，所以半径也是 $a$。再把点 $(5,6)$ 代入圆方程即可求出 $a$。

Let the centre of the circle be $(a,0)$, where $a>0$.

Since the circle touches the $y$-axis, its radius is $a$.

Therefore the equation of $C$ is

Since the circle passes through $(5,6)$,

Hence

也可以直接设半径为 $r$。从圆心到点 $(5,6)$ 的水平距离是 $r-5$，竖直距离是 $6$，再用 Pythagoras’ theorem。

Let the radius of the circle be $r$.

Since the circle touches the $y$-axis and has centre on the $x$-axis, its centre is $(r,0)$.

The point $(5,6)$ lies on the circle, so the distance from $(r,0)$ to $(5,6)$ is $r$:

Therefore the centre is $\left(\frac{61}{10},0\right)$ and the radius is $\frac{61}{10}$.

So

圆经过原点和 $(5,6)$。连接这两个点是一条 chord，圆心必在这条 chord 的 perpendicular bisector 上。再与 $x$-axis 相交，就能找到圆心。

Since the circle touches the $y$-axis at the origin, the points $(0,0)$ and $(5,6)$ both lie on the circle.

The midpoint of the chord joining these points is

The gradient of the chord is

So the gradient of the perpendicular bisector is

The perpendicular bisector is

The centre lies on the $x$-axis, so set $y=0$:

Therefore the centre is $\left(\frac{61}{10},0\right)$.

The radius is also $\frac{61}{10}$, since the circle touches the $y$-axis at the origin.

Hence