Circle Tangent B722D0

1. **State the problem:** We have a circle with equation $$(x-6)^2 + (y+a)^2 = 18$$ and a line $$y = 2a - x$$ that is tangent to the circle. We need to find the values of $$a$$ for which the line is tangent, and then find the equation of a diameter perpendicular to this tangent for the greater $$a$$. 2. **Recall the tangent condition:** A line is tangent to a circle if the distance from the center of the circle to the line equals the radius. 3. **Identify the center and radius:** Center: $$(6, -a)$$ Radius: $$r = \sqrt{18} = 3\sqrt{2}$$ 4. **Rewrite the line in standard form:** Given $$y = 2a - x$$, rearranged: $$x + y - 2a = 0$$ 5. **Distance from center to line formula:** $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ where $$A=1$$, $$B=1$$, $$C=-2a$$, and center $$(x_0,y_0) = (6,-a)$$. 6. **Calculate the distance:** $$d = \frac{|1 \cdot 6 + 1 \cdot (-a) - 2a|}{\sqrt{1^2 + 1^2}} = \frac{|6 - a - 2a|}{\sqrt{2}} = \frac{|6 - 3a|}{\sqrt{2}}$$ 7. **Set distance equal to radius:** $$\frac{|6 - 3a|}{\sqrt{2}} = 3\sqrt{2}$$ 8. **Solve for $$a$$:** Multiply both sides by $$\sqrt{2}$$: $$|6 - 3a| = 3\sqrt{2} \times \sqrt{2} = 3 \times 2 = 6$$ So, $$|6 - 3a| = 6$$ This gives two cases: - Case 1: $$6 - 3a = 6 \Rightarrow -3a = 0 \Rightarrow a = 0$$ - Case 2: $$6 - 3a = -6 \Rightarrow -3a = -12 \Rightarrow a = 4$$ **Thus, the two possible values of $$a$$ are $$0$$ and $$4$$.** 9. **For part (b), use the greater value $$a=4$$.** 10. **Find the slope of the tangent line:** Given $$y = 2a - x$$, slope $$m = -1$$. 11. **Slope of diameter perpendicular to tangent:** Perpendicular slope $$m_{\perp} = 1$$ (negative reciprocal of $$-1$$). 12. **Equation of diameter passing through center $$(6, -a) = (6, -4)$$:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - (-4) = 1(x - 6)$$ $$y + 4 = x - 6$$ $$y = x - 10$$ **Final answers:** (a) $$a = 0$$ or $$a = 4$$. (b) For $$a=4$$, the diameter equation is $$y = x - 10$$.