1. **Problem:** We need to find the coordinates of the point where the two circles touch. The circles are given by the equations $$((x+7)^2)+((y+2)^2)=4$$ and $$((x+7)^2)+((y+5)^2)=16$$. 2. **Identify centers and radii:** The first circle has center $$C_1=(-7,-2)$$ and radius $$r_1=2$$ (since $$4=2^2$$). The second circle has center $$C_2=(-7,-5)$$ and radius $$r_2=4$$ (since $$16=4^2$$). 3. **Check the distance between centers:** Calculate the distance \\(d\\) between centers $$C_1$$ and $$C_2$$ by $$d=\sqrt{(-7+7)^2+(-2+5)^2}=\sqrt{0^2 + 3^2}=3$$. 4. **Check if circles touch externally:** Two circles touch externally if the distance between centers equals the sum of the radii: $$d = r_1 + r_2\\$$ Here, $$3 = 2 + 4 = 6$$ which is false. 5. **Check if circles touch internally:** Two circles touch internally if the distance between centers equals the absolute difference of the radii: $$d = |r_2 - r_1|\\$$ Here, $$3 = |4 - 2| = 2$$ which is false. 6. **Re-examine the problem:** Actually, distance between centers $$d=3$$, radii $$r_1=2$$ and $$r_2=4$$, so the circles intersect in two points if $$|r_2 - r_1| < d < r_1 + r_2\\$$ which is $$2 < 3 < 6$$ true, meaning they intersect at two points. 7. **However, the problem states they touch at one point,** so likely one radius is incorrectly assigned; check carefully. 8. **Recalculate carefully:** Centers: $$C_1=(-7,-2), r_1=2$$ $$C_2=(-7,-5), r_2=4$$ Distance between centers: $$d=3$$ Sum of radii: $$r_1 + r_2 = 6$$ Difference of radii: $$|r_2 - r_1| = 2$$ Since $$d=3$$ is between 2 and 6, they intersect in two points, so must find intersection points. 9. **Find intersection points:** Set the two equations equal: $$((x+7)^2)+((y+2)^2)=4$$ $$((x+7)^2)+((y+5)^2)=16$$ Subtract first from second to eliminate $x$: $$((y+5)^2) - ((y+2)^2) = 12$$ Simplify: $$(y+5)^2 - (y+2)^2 = (y^2 + 10y + 25) - (y^2 + 4y + 4) = 6y + 21 = 12$$ So $$6y = 12 - 21 = -9\\$$ $$y = -\frac{9}{6} = -\frac{3}{2} = -1.5$$ 10. **Substitute $$y = -1.5$$ into first circle:** $$((x+7)^2) + (-1.5 + 2)^2 = 4$$ $$((x+7)^2) + (0.5)^2 = 4$$ $$((x+7)^2) + 0.25 = 4$$ $$((x+7)^2) = 3.75$$ $$x + 7 = \pm \sqrt{3.75} = \pm 1.936492$$ 11. **Two possible x coordinates:** $$x = -7 + 1.936492 = -5.063508$$ $$x = -7 - 1.936492 = -8.936492$$ 12. **Check which point also satisfies the second circle:** For $$x=-5.063508, y=-1.5$$: $$((x+7)^2) + ((y+5)^2) = (1.936492)^2 + (3.5)^2 = 3.75 + 12.25 = 16$$ satisfies second circle. Similarly for $$x=-8.936492$$ also true. Thus, the circles intersect at two points and do not touch at only one point, so the problem's claim about one touching point is incorrect. **However, given the question asks for the touching point, likely a misprint, maybe the radius of the second circle is incorrect; if instead the second circle's radius is 3, i.e., $$((x+7)^2) + ((y+5)^2) = 9$$ then $$d = 3$$ equals $$r_1 + r_2 = 2 + 3 = 5$$ which is false, but if radius 1 is 2 and radius 2 is 1, difference equals 1 and distance 3, no touching. Alternatively, if difference of radii is 3 (4-1=3), distance 3, then circles touch internally at one point. Check for $$((x+7)^2) + ((y+5)^2) = 1$$ radius 1 with radius 2: $$r_1 = 2,$$ $$r_2 =1$$ $$|r_1 - r_2| = 1$$, distance 3 no, Conclusion: **The circles touch at exactly one point if the distance between centers equals the sum or difference of radii, here it equals neither, so the circles intersect in two points, not touch at one.** **Therefore, the point(s) of intersection are:** $$(-5.06, -1.5)$$ and $$(-8.94, -1.5)$$ (rounded to two decimals).