1. **Problem statement:** Find the equation of the circle passing through points $(1,2)$ and $(3,4)$ and tangent to the line $3x + y - 3 = 0$. 2. **General form of a circle:** The equation of a circle can be written as $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $(h,k)$ is the center and $r$ is the radius. 3. **Using the points:** Since the circle passes through $(1,2)$ and $(3,4)$, these points satisfy the circle equation: $$ (1 - h)^2 + (2 - k)^2 = r^2 $$ $$ (3 - h)^2 + (4 - k)^2 = r^2 $$ 4. **Equating the two expressions for $r^2$:** $$ (1 - h)^2 + (2 - k)^2 = (3 - h)^2 + (4 - k)^2 $$ Expanding and simplifying: $$ (1 - h)^2 - (3 - h)^2 + (2 - k)^2 - (4 - k)^2 = 0 $$ Using the identity $a^2 - b^2 = (a-b)(a+b)$: $$ [(1 - h) - (3 - h)] imes [(1 - h) + (3 - h)] + [(2 - k) - (4 - k)] imes [(2 - k) + (4 - k)] = 0 $$ Simplify each term: $$ (1 - h - 3 + h)(1 - h + 3 - h) + (2 - k - 4 + k)(2 - k + 4 - k) = 0 $$ $$ (-2)(4 - 2h) + (-2)(6 - 2k) = 0 $$ $$ -8 + 4h -12 + 4k = 0 $$ $$ 4h + 4k - 20 = 0 $$ Divide by 4: $$ h + k = 5 $$ 5. **Center lies on the line:** $$ k = 5 - h $$ 6. **Radius and tangent condition:** The radius $r$ is the distance from the center $(h,k)$ to either point, for example $(1,2)$: $$ r = \sqrt{(1 - h)^2 + (2 - k)^2} $$ 7. **Tangency condition:** The distance from the center $(h,k)$ to the tangent line $3x + y - 3 = 0$ equals the radius $r$: $$ \text{Distance} = \frac{|3h + k - 3|}{\sqrt{3^2 + 1^2}} = r $$ $$ \Rightarrow \frac{|3h + k - 3|}{\sqrt{10}} = r $$ 8. **Equate radius expressions:** $$ \sqrt{(1 - h)^2 + (2 - k)^2} = \frac{|3h + k - 3|}{\sqrt{10}} $$ Square both sides: $$ (1 - h)^2 + (2 - k)^2 = \frac{(3h + k - 3)^2}{10} $$ 9. **Substitute $k = 5 - h$:** $$ (1 - h)^2 + (2 - (5 - h))^2 = \frac{(3h + (5 - h) - 3)^2}{10} $$ Simplify inside: $$ (1 - h)^2 + (2 - 5 + h)^2 = \frac{(3h + 5 - h - 3)^2}{10} $$ $$ (1 - h)^2 + (-3 + h)^2 = \frac{(2h + 2)^2}{10} $$ 10. **Expand and simplify:** $$ (1 - h)^2 = (h - 1)^2 = h^2 - 2h + 1 $$ $$ (-3 + h)^2 = (h - 3)^2 = h^2 - 6h + 9 $$ $$ (2h + 2)^2 = 4(h + 1)^2 = 4(h^2 + 2h + 1) = 4h^2 + 8h + 4 $$ Sum left side: $$ h^2 - 2h + 1 + h^2 - 6h + 9 = 2h^2 - 8h + 10 $$ Right side: $$ \frac{4h^2 + 8h + 4}{10} = \frac{2h^2 + 4h + 2}{5} $$ 11. **Multiply both sides by 10 to clear denominator:** $$ 10(2h^2 - 8h + 10) = 4h^2 + 8h + 4 $$ $$ 20h^2 - 80h + 100 = 4h^2 + 8h + 4 $$ 12. **Bring all terms to one side:** $$ 20h^2 - 80h + 100 - 4h^2 - 8h - 4 = 0 $$ $$ 16h^2 - 88h + 96 = 0 $$ Divide by 4: $$ 4h^2 - 22h + 24 = 0 $$ 13. **Solve quadratic:** $$ h = \frac{22 \pm \sqrt{(-22)^2 - 4 \times 4 \times 24}}{2 \times 4} = \frac{22 \pm \sqrt{484 - 384}}{8} = \frac{22 \pm \sqrt{100}}{8} $$ $$ h = \frac{22 \pm 10}{8} $$ Two solutions: - $$ h = \frac{22 + 10}{8} = \frac{32}{8} = 4 $$ - $$ h = \frac{22 - 10}{8} = \frac{12}{8} = 1.5 $$ 14. **Find corresponding $k$ values:** - For $h=4$, $$ k = 5 - 4 = 1 $$ - For $h=1.5$, $$ k = 5 - 1.5 = 3.5 $$ 15. **Find radius $r$ for each center:** - For $(4,1)$: $$ r = \sqrt{(1 - 4)^2 + (2 - 1)^2} = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} $$ - For $(1.5, 3.5)$: $$ r = \sqrt{(1 - 1.5)^2 + (2 - 3.5)^2} = \sqrt{(-0.5)^2 + (-1.5)^2} = \sqrt{0.25 + 2.25} = \sqrt{2.5} $$ 16. **Write the equations of the circles:** - For center $(4,1)$ and radius $\sqrt{10}$: $$ (x - 4)^2 + (y - 1)^2 = 10 $$ - For center $(1.5, 3.5)$ and radius $\sqrt{2.5}$: $$ (x - 1.5)^2 + (y - 3.5)^2 = 2.5 $$ **Final answer:** The two possible circle equations are: $$ (x - 4)^2 + (y - 1)^2 = 10 $$ $$ (x - 1.5)^2 + (y - 3.5)^2 = 2.5 $$