Animated Solution for Mathematics - Circles: Let a circle $C:(x-h{)}^2+(y-k{)}^2=r^2,k>0$, touch the x-axis at $(1,0)$. If the line $x+y=0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $PQ$ is 2, then the value of $h+k+r$ is equal to ____.

The circle $C$ is tangent to the $x$-axis at the point $(1,0)$. This geometric constraint implies that the center of the circle must lie on the vertical line $x=1$.

Let the center of the circle be $(h,k)$. From the tangency condition, we immediately identify $h=1$. Furthermore, the radius $r$ must be equal to the vertical distance from the center to the $x$-axis, so $r=k$.

The equation of the circle can therefore be expressed as:

We are given that the line $x+y=0$ intersects the circle to form a chord $PQ$ of length $2$. Instead of solving for intersection points, we utilize the relationship between the chord length $L$, the radius $r$, and the perpendicular distance $d$ from the center to the line:

Given $L=2$, we simplify this to $1=\sqrt{r^2-d^2}$, or $1=r^2-d^2$. Substituting $r=k$, we have $1=k^2-d^2$.

The perpendicular distance $d$ from the center $(1,k)$ to the line $x+y=0$ is calculated using the point-to-line distance formula:

Since the circle must exist in the upper half-plane for the tangency condition to hold with $k>0$, we simplify this to:

Substituting the expression for $d$ into our chord equation $1=k^2-d^2$:

Multiplying the entire equation by $2$ to clear the denominator:

Factoring the quadratic equation, we obtain:

This yields roots $k=3$ and $k=-1$. Given the constraint $k>0$, we discard $k=-1$ and accept $k=3$.

Since $h=1$, $k=3$, and $r=k=3$, the sum of these values is: