Circle Equation 118E8A

1. **State the problem:** Find the equation of a circle passing through the intersection points of two given circles and tangent to a given line. 2. **Given circles:** $$C_1: x^2 + y^2 - 4 = 0$$ $$C_2: x^2 + y^2 - 2x - 4y + 4 = 0$$ 3. **General form of the required circle:** The required circle passes through the intersection of $C_1$ and $C_2$, so its equation can be written as: $$C = C_1 + \lambda (C_2) = 0$$ which means: $$x^2 + y^2 - 4 + \lambda (x^2 + y^2 - 2x - 4y + 4) = 0$$ 4. **Expand and group terms:** $$x^2 + y^2 - 4 + \lambda x^2 + \lambda y^2 - 2\lambda x - 4\lambda y + 4\lambda = 0$$ $$\Rightarrow (1 + \lambda) x^2 + (1 + \lambda) y^2 - 2\lambda x - 4\lambda y - 4 + 4\lambda = 0$$ 5. **Divide entire equation by $(1 + \lambda)$ (assuming $1 + \lambda \neq 0$):** $$x^2 + y^2 - \frac{2\lambda}{1 + \lambda} x - \frac{4\lambda}{1 + \lambda} y - \frac{4 - 4\lambda}{1 + \lambda} = 0$$ Intermediate step showing cancellation: $$x^2 + y^2 - \cancel{\frac{2\lambda}{1 + \lambda}} x - \cancel{\frac{4\lambda}{1 + \lambda}} y - \cancel{\frac{4 - 4\lambda}{1 + \lambda}} = 0$$ 6. **Rewrite in standard circle form:** Center: $$\left(\frac{\lambda}{1 + \lambda}, \frac{2\lambda}{1 + \lambda}\right)$$ Radius squared: $$r^2 = \left(\frac{\lambda}{1 + \lambda}\right)^2 + \left(\frac{2\lambda}{1 + \lambda}\right)^2 + \frac{4 - 4\lambda}{1 + \lambda}$$ 7. **Condition for tangency to line $x + 2y = 0$:** Distance from center to line equals radius: $$\frac{|h + 2k|}{\sqrt{1^2 + 2^2}} = r$$ where $h = \frac{\lambda}{1 + \lambda}$ and $k = \frac{2\lambda}{1 + \lambda}$. Calculate numerator: $$|h + 2k| = \left| \frac{\lambda}{1 + \lambda} + 2 \cdot \frac{2\lambda}{1 + \lambda} \right| = \left| \frac{\lambda + 4\lambda}{1 + \lambda} \right| = \left| \frac{5\lambda}{1 + \lambda} \right|$$ Denominator: $$\sqrt{1 + 4} = \sqrt{5}$$ So, $$\frac{|5\lambda/(1 + \lambda)|}{\sqrt{5}} = r$$ 8. **Square both sides:** $$\frac{25 \lambda^2}{(1 + \lambda)^2 \cdot 5} = r^2$$ $$\Rightarrow \frac{5 \lambda^2}{(1 + \lambda)^2} = r^2$$ 9. **Equate radius squared expressions:** $$r^2 = \left(\frac{\lambda}{1 + \lambda}\right)^2 + \left(\frac{2\lambda}{1 + \lambda}\right)^2 + \frac{4 - 4\lambda}{1 + \lambda}$$ $$= \frac{\lambda^2}{(1 + \lambda)^2} + \frac{4 \lambda^2}{(1 + \lambda)^2} + \frac{4 - 4\lambda}{1 + \lambda}$$ $$= \frac{5 \lambda^2}{(1 + \lambda)^2} + \frac{4 - 4\lambda}{1 + \lambda}$$ Set equal to previous expression: $$\frac{5 \lambda^2}{(1 + \lambda)^2} = \frac{5 \lambda^2}{(1 + \lambda)^2} + \frac{4 - 4\lambda}{1 + \lambda}$$ Subtract $\frac{5 \lambda^2}{(1 + \lambda)^2}$ from both sides: $$0 = \frac{4 - 4\lambda}{1 + \lambda}$$ 10. **Solve for $\lambda$:** $$4 - 4\lambda = 0$$ $$\Rightarrow 4\lambda = 4$$ $$\Rightarrow \lambda = 1$$ 11. **Substitute $\lambda = 1$ back into circle equation:** $$C = C_1 + 1 \cdot C_2 = 0$$ $$x^2 + y^2 - 4 + x^2 + y^2 - 2x - 4y + 4 = 0$$ $$2x^2 + 2y^2 - 2x - 4y = 0$$ Divide entire equation by 2: $$x^2 + y^2 - x - 2y = 0$$ 12. **Rewrite in standard form:** Complete the square: $$x^2 - x + y^2 - 2y = 0$$ $$x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + y^2 - 2y + 1 - 1 = 0$$ $$\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - 1\right)^2 - 1 = 0$$ $$\left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = \frac{5}{4}$$ **Final answer:** $$\boxed{\left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = \frac{5}{4}}$$ This is the equation of the required circle.