1. **Problem:** Write down the centre and radius length of the circle $$x^2 + y^2 - 6x - 2y - 15 = 0$$. If the line $$3x + 4y + c = 0$$ is a tangent to this circle, find the two values of $$c$$. 2. **Step 1: Find the centre and radius of the circle.** The general form of a circle is $$(x - h)^2 + (y - k)^2 = r^2$$ where $$(h, k)$$ is the centre and $$r$$ is the radius. Rewrite the given equation by completing the square: $$x^2 - 6x + y^2 - 2y = 15$$ Complete the square for $$x$$: $$x^2 - 6x = (x^2 - 6x + 9) - 9 = (x - 3)^2 - 9$$ Complete the square for $$y$$: $$y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1$$ Substitute back: $$ (x - 3)^2 - 9 + (y - 1)^2 - 1 = 15 $$ Simplify: $$ (x - 3)^2 + (y - 1)^2 - 10 = 15 $$ $$ (x - 3)^2 + (y - 1)^2 = 25 $$ So, the centre is $$C = (3, 1)$$ and the radius is $$r = 5$$. 3. **Step 2: Condition for tangency of a line to a circle.** A line $$Ax + By + C = 0$$ is tangent to a circle with centre $$(h, k)$$ and radius $$r$$ if the perpendicular distance from the centre to the line equals the radius: $$\text{Distance} = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} = r$$ 4. **Step 3: Apply the condition to the line $$3x + 4y + c = 0$$.** Here, $$A = 3$$, $$B = 4$$, and $$C = c$$. Calculate the distance from centre $$(3, 1)$$ to the line: $$\frac{|3(3) + 4(1) + c|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 4 + c|}{5} = \frac{|13 + c|}{5}$$ Set equal to radius $$5$$: $$\frac{|13 + c|}{5} = 5$$ Multiply both sides by 5: $$|13 + c| = 25$$ 5. **Step 4: Solve for $$c$$.** $$13 + c = 25 \quad \text{or} \quad 13 + c = -25$$ $$c = 25 - 13 = 12 \quad \text{or} \quad c = -25 - 13 = -38$$ **Final answers:** - Centre: $$ (3, 1) $$ - Radius: $$ 5 $$ - Values of $$c$$ for tangency: $$12$$ and $$-38$$