1. **Stating the problem:** We need to find the equation of a circle $k$ centered in the first quadrant. Given: - The circle is tangent to the y-axis at point $(0,3)$. - The circle intersects the x-axis at points $G$ and $H$. - The length of segment $GH$ on the x-axis is 8. 2. **Formula for a circle:** The general equation of a circle with center $(h,k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ 3. **Using the tangent condition:** Since the y-axis ($x=0$) is tangent to the circle at $(0,3)$, the distance from the center to the y-axis equals the radius. - The distance from center $(h,k)$ to y-axis is $|h|$. - The radius $r$ equals $|h|$. Also, the point $(0,3)$ lies on the circle, so: $$ (0 - h)^2 + (3 - k)^2 = r^2 $$ Substitute $r = h$ (since $h > 0$ in the first quadrant): $$ h^2 + (3 - k)^2 = h^2 $$ Simplify: $$ \cancel{h^2} + (3 - k)^2 = \cancel{h^2} $$ $$ (3 - k)^2 = 0 $$ $$ 3 - k = 0 \Rightarrow k = 3 $$ 4. **Center coordinates:** The center is $(h,3)$ with $h > 0$. 5. **Using the intersection with x-axis:** The circle intersects the x-axis where $y=0$. Substitute $y=0$ into the circle equation: $$ (x - h)^2 + (0 - 3)^2 = r^2 $$ Recall $r = h$, so: $$ (x - h)^2 + 9 = h^2 $$ Simplify: $$ (x - h)^2 = h^2 - 9 $$ 6. **Finding points $G$ and $H$ on x-axis:** The solutions for $x$ are: $$ x - h = \pm \sqrt{h^2 - 9} $$ $$ x = h \pm \sqrt{h^2 - 9} $$ 7. **Length of segment $GH$:** $$ |GH| = \left| (h + \sqrt{h^2 - 9}) - (h - \sqrt{h^2 - 9}) \right| = 2 \sqrt{h^2 - 9} $$ Given $|GH| = 8$, so: $$ 2 \sqrt{h^2 - 9} = 8 $$ Divide both sides by 2: $$ \cancel{2} \sqrt{h^2 - 9} = \cancel{2} 4 $$ $$ \sqrt{h^2 - 9} = 4 $$ Square both sides: $$ h^2 - 9 = 16 $$ $$ h^2 = 25 $$ Since $h > 0$ (first quadrant), $$ h = 5 $$ 8. **Radius:** $$ r = h = 5 $$ 9. **Equation of the circle:** Center: $(5,3)$, radius: $5$ $$ (x - 5)^2 + (y - 3)^2 = 25 $$ **Final answer:** The equation of the circle $k$ is $$ (x - 5)^2 + (y - 3)^2 = 25 $$