1. **Problem statement:** Find the coordinates of point B where the circle centered at O intersects the x-axis, and find the equation of the line on which the diameter BD lies. 2. **Given:** The chord AB lies on the line $y = x + 4$. 3. **Step 1: Find coordinates of B on the x-axis.** - Since B lies on the x-axis, its $y$-coordinate is 0. - Substitute $y=0$ into the line equation $y = x + 4$: $$0 = x + 4 \implies x = -4$$ - So, $B = (-4, 0)$. 4. **Step 2: Find coordinates of A.** - Point A lies on the line $y = x + 4$ and on the circle centered at O. - Since the circle is centered at O (origin), assume radius $r$. - The circle equation is $x^2 + y^2 = r^2$. - On line $y = x + 4$, substitute $y$: $$x^2 + (x+4)^2 = r^2$$ - Expand: $$x^2 + x^2 + 8x + 16 = r^2$$ $$2x^2 + 8x + 16 = r^2$$ 5. **Step 3: Use point B to find $r^2$.** - Point B is on the circle, so: $$(-4)^2 + 0^2 = r^2 \implies 16 = r^2$$ 6. **Step 4: Solve for $x$ of point A.** - Substitute $r^2=16$: $$2x^2 + 8x + 16 = 16$$ $$2x^2 + 8x = 0$$ $$2x(x + 4) = 0$$ - Solutions: $$x=0 \text{ or } x=-4$$ - $x=-4$ corresponds to point B, so $x=0$ is point A. - Find $y$: $$y = 0 + 4 = 4$$ - So, $A = (0, 4)$. 7. **Step 5: Equation of the circle.** - Center at $O(0,0)$, radius $r=4$. - Equation: $$x^2 + y^2 = 16$$ 8. **Step 6: Find coordinates of D on the circle in the third quadrant.** - Since D lies in the third quadrant, $x<0$, $y<0$. - Let $D = (x_D, y_D)$ satisfy: $$x_D^2 + y_D^2 = 16$$ 9. **Step 7: Equation of diameter BD.** - Diameter BD passes through points $B(-4,0)$ and $D(x_D,y_D)$. - The midpoint of BD is the center $O(0,0)$: $$\left(\frac{-4 + x_D}{2}, \frac{0 + y_D}{2}\right) = (0,0)$$ - Solve for $x_D$ and $y_D$: $$\frac{-4 + x_D}{2} = 0 \implies x_D = 4$$ $$\frac{0 + y_D}{2} = 0 \implies y_D = 0$$ - But $D$ must be in the third quadrant, so this contradicts. 10. **Step 8: Reconsider the midpoint condition.** - Since BD is a diameter, midpoint is center $O$. - Given $B(-4,0)$, midpoint $O(0,0)$, then: $$x_D = -x_B = 4$$ $$y_D = -y_B = 0$$ - So $D = (4,0)$ lies on the positive x-axis, not third quadrant. 11. **Step 9: Conclusion about D.** - The problem states D is in the third quadrant on the circle. - Since B is at $(-4,0)$, the point diametrically opposite is $(4,0)$. - To have D in the third quadrant, D must be different from the diameter endpoint opposite B. - Possibly, the diameter BD is not the diameter through O, or the problem needs more data. 12. **Step 10: Equation of line BD.** - Using points $B(-4,0)$ and $D(x_D,y_D)$, slope: $$m = \frac{y_D - 0}{x_D + 4}$$ - Equation: $$y - 0 = m(x + 4)$$ **Final answers:** - Coordinates of $B$ are $(-4,0)$. - Equation of the circle: $$x^2 + y^2 = 16$$. - Equation of line $AB$: $$y = x + 4$$. - Diameter BD passes through $B(-4,0)$ and $D(4,0)$. - Equation of diameter BD: $$y = 0$$ (the x-axis). Note: The problem's statement about D in the third quadrant contradicts the diameter definition with B at $(-4,0)$ and center at origin. More information is needed to resolve this.