Let C_1$C_1$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let C_2$C_2$ be the circle with centre (1, 3) that touches C_1$C_1$ externally at the point (alpha, beta)$(\alpha, \beta)$. If (beta - alpha)^2 = fracmn$(\beta - \alpha)^2 = \frac{m}{n}$, gcd(m, n) = 1$\gcd(m, n) = 1$, then m + n$m + n$ is equal to:

### Related Formula For a circle in the third quadrant touching both coordinate axes, the center layout is (-r, -r)$(-r, -r)$ and equation looks like: (x + r)^2 + (y + r)^2 = r^2$$(x + r)^2 + (y + r)^2 = r^2$$ For external contact between circles C_1$C_1$ and C_2$C_2$, the distance between centers equals the sum of their radii: C_1C_2 = r_1 + r_2$$C_1C_2 = r_1 + r_2$$ ### Core Logic Circle C_1$C_1$ has radius r_1 = 3$r_1 = 3$ and touches both axes in the third quadrant, so its center is A(-3, -3)$A(-3, -3)$. Circle C_2$C_2$ has center B(1, 3)$B(1, 3)$. The distance between centers A$A$ and B$B$ is: AB = sqrt(1 - (-3))^2 + (3 - (-3))^2 = sqrt4^2 + 6^2 = sqrt16 + 36 = sqrt52 = 2sqrt13$$AB = \sqrt{(1 - (-3))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$ ### Step 1: Determine Radius of Circle 2

Since the circles touch externally: AB = r_1 + r_2 implies 2sqrt13 = 3 + r_2 implies r_2 = 2sqrt13 - 3$$AB = r_1 + r_2 \implies 2\sqrt{13} = 3 + r_2 \implies r_2 = 2\sqrt{13} - 3$$ ### Step 2: Locate the Contact Point via Section Formula The point of contact P(alpha, beta)$P(\alpha, \beta)$ divides the line segment joining centers A(-3, -3)$A(-3, -3)$ and B(1, 3)$B(1, 3)$ internally in the ratio r_1 : r_2 = 3 : (2sqrt13 - 3)$r_1 : r_2 = 3 : (2\sqrt{13} - 3)$. Using the internal section formula: alpha = frac3(1) + (2sqrt13 - 3)(-3)3 + (2sqrt13 - 3) = frac3 - 6sqrt13 + 92sqrt13 = frac12 - 6sqrt13 + 02sqrt13 = frac6 - 3sqrt13sqrt13$$\alpha = \frac{3(1) + (2\sqrt{13} - 3)(-3)}{3 + (2\sqrt{13} - 3)} = \frac{3 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{12 - 6\sqrt{13} + 0}{2\sqrt{13}} = \frac{6 - 3\sqrt{13}}{\sqrt{13}}$$ beta = frac3(3) + (2sqrt13 - 3)(-3)3 + (2sqrt13 - 3) = frac9 - 6sqrt13 + 92sqrt13 = frac18 - 6sqrt132sqrt13 = frac9 - 3sqrt13sqrt13$$\beta = \frac{3(3) + (2\sqrt{13} - 3)(-3)}{3 + (2\sqrt{13} - 3)} = \frac{9 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}} = \frac{9 - 3\sqrt{13}}{\sqrt{13}}$$ ### Step 3: Calculate the Difference Value Find (beta - alpha)^2$(\beta - \alpha)^2$: beta - alpha = frac9 - 3sqrt13sqrt13 - frac6 - 3sqrt13sqrt13 = frac3sqrt13$$\beta - \alpha = \frac{9 - 3\sqrt{13}}{\sqrt{13}} - \frac{6 - 3\sqrt{13}}{\sqrt{13}} = \frac{3}{\sqrt{13}}$$ (beta - alpha)^2 = left(frac3sqrt13right)^2 = frac913$$(\beta - \alpha)^2 = \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13}$$ Comparing with fracmn$\frac{m}{n}$ where gcd(m, n) = 1$\gcd(m, n) = 1$ gives m = 9, n = 13$m = 9, n = 13$. m + n = 9 + 13 = 22$$m + n = 9 + 13 = 22$$ ### Pattern Recognition Notice that computing (beta - alpha)$(\beta - \alpha)$ directly cancels out the irrational sqrt13$\sqrt{13}$ term from the numerator before squaring, saving a significant amount of tedious arithmetic expansion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Coordinate Geometry Class 11 Mathematics: Circles