Let the set of all values of r$r$, for which the circles (x + 1)^2 + (y + 4)^2 = r^2$(x + 1)^2 + (y + 4)^2 = r^2$ and x^2 + y^2 - 4x - 2y - 4 = 0$x^2 + y^2 - 4x - 2y - 4 = 0$ intersect at two distinct points be the interval (alpha, beta)$(\alpha, \beta)$. Then alphabeta$\alpha\beta$ is equal to

### Related Formula For external tangents from point R$R$ forming angle 2theta$2\theta$ at the center, the distance d$d$ from center to intersection point obeys cos theta = fracrd$\cos \theta = \frac{r}{d}$. ### Core Logic Given circle: x^2 + y^2 - 4x - 6y - 3 = 0$x^2 + y^2 - 4x - 6y - 3 = 0$ Center O = (2, 3)$O = (2, 3)$ Radius r = sqrt(-2)^2 + (-3)^2 - (-3) = sqrt4 + 9 + 3 = sqrt16 = 4$r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4$. The tangents PQ and MN intersect at some point R(h, k)$R(h, k)$. The radius vectors OA$OA$ and OB$OB$ subtend angle angle AOB = fracpi3 = 60^circ$\angle AOB = \frac{\pi}{3} = 60^{\circ}$ at the center. The line joining O$O$ and R$R$ bisects the angle angle AOB$\angle AOB$. Thus, angle AOR = 30^circ$\angle AOR = 30^{\circ}$. ### Step 1: Apply Trigonometric Relations

In the right-angled triangle Delta AOR$\Delta AOR$, OA$OA$ is the radius (r = 4$r = 4$) and OR$OR$ is the hypotenuse. cos(30^circ) = fracOAOR = fracrOR$$\cos(30^{\circ}) = \frac{OA}{OR} = \frac{r}{OR}$$ fracsqrt32 = frac4OR Rightarrow OR = frac8sqrt3$$\frac{\sqrt{3}}{2} = \frac{4}{OR} \Rightarrow OR = \frac{8}{\sqrt{3}}$$ ### Step 2: Construct the Locus Equation The distance squared between O(2,3)$O(2,3)$ and R(h,k)$R(h,k)$ is OR^2$OR^2$: OR^2 = (h - 2)^2 + (k - 3)^2 = left(frac8sqrt3right)^2$$OR^2 = (h - 2)^2 + (k - 3)^2 = \left(\frac{8}{\sqrt{3}}\right)^2$$ (h - 2)^2 + (k - 3)^2 = frac643$$(h - 2)^2 + (k - 3)^2 = \frac{64}{3}$$ h^2 - 4h + 4 + k^2 - 6k + 9 = frac643$$h^2 - 4h + 4 + k^2 - 6k + 9 = \frac{64}{3}$$ 3(h^2 + k^2 - 4h - 6k + 13) = 64$$3(h^2 + k^2 - 4h - 6k + 13) = 64$$ 3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$ 3(h^2 + k^2) - 12h - 18k - 25 = 0$$3(h^2 + k^2) - 12h - 18k - 25 = 0$$ ### Step 3: Generalize the Equation Replace (h, k)$(h, k)$ with (x, y)$(x, y)$ for the general locus: 3(x^2 + y^2) - 12x - 18y - 25 = 0$$3(x^2 + y^2) - 12x - 18y - 25 = 0$$ ### Pattern Recognition The locus of the intersection of tangents enclosing a constant angle is simply a concentric circle. Its radius expands by 1/sin(alpha/2)$1/\sin(\alpha/2)$ or 1/cos(theta)$1/\cos(\theta)$ depending on whether the angle is measured at intersection or center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

### Related Formula textPerpendicular bisector properties: m_1m_2 = -1 text and mid-point lies on the line.$$\text{Perpendicular bisector properties: } m_1m_2 = -1 \text{ and mid-point lies on the line.}$$ textParametric point on circle x^2+y^2=r^2 text is (rcostheta, rsintheta)$$\text{Parametric point on circle } x^2+y^2=r^2 \text{ is } (r\cos\theta, r\sin\theta)$$ ### Core Logic

Let P = (2costheta, 2sintheta)$P = (2\cos\theta, 2\sin\theta)$. Let Q$Q$ on the line 5x + y + 2 = 0$5x + y + 2 = 0$ be Q(alpha, -5alpha-2)$Q(\alpha, -5\alpha-2)$. The line x - y + 1 = 0$x - y + 1 = 0$ is the perpendicular bisector of PQ$PQ$. This gives two conditions: slope of PQ$PQ$ is -1$-1$, and mid-point of PQ$PQ$ satisfies the bisector equation. ### Step 1: Apply Slope Condition Slope of bisector is 1$1$, so slope of PQ$PQ$ must be -1$-1$. frac2sintheta - (-5alpha - 2)2costheta - alpha = -1$$\frac{2\sin\theta - (-5\alpha - 2)}{2\cos\theta - \alpha} = -1$$ 2sintheta + 5alpha + 2 = -2costheta + alpha$$2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$$ sintheta + costheta + 2alpha + 1 = 0 quad dots (1)$$\sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad \dots (1)$$ ### Step 2: Apply Midpoint Condition Midpoint of PQ$PQ$ is left( frac2costheta + alpha2, frac2sintheta - 5alpha - 22 right)$\left( \frac{2\cos\theta + \alpha}{2}, \frac{2\sin\theta - 5\alpha - 2}{2} \right)$. Substitute into x - y + 1 = 0$x - y + 1 = 0$: frac2costheta + alpha2 - frac2sintheta - 5alpha - 22 + 1 = 0$$\frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$$ 2costheta + alpha - 2sintheta + 5alpha + 2 + 2 = 0$$2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$$ costheta - sintheta + 3alpha + 2 = 0 quad dots (2)$$\cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad \dots (2)$$ ### Step 3: Eliminate alpha and Solve From (1), 2alpha = -sintheta - costheta - 1 implies alpha = frac-sintheta - costheta - 12$2\alpha = -\sin\theta - \cos\theta - 1 \implies \alpha = \frac{-\sin\theta - \cos\theta - 1}{2}$. Substitute alpha$\alpha$ into (2): costheta - sintheta + 3left( frac-sintheta - costheta - 12 right) + 2 = 0$$\cos\theta - \sin\theta + 3\left( \frac{-\sin\theta - \cos\theta - 1}{2} \right) + 2 = 0$$ 2costheta - 2sintheta - 3sintheta - 3costheta - 3 + 4 = 0$$2\cos\theta - 2\sin\theta - 3\sin\theta - 3\cos\theta - 3 + 4 = 0$$ -costheta - 5sintheta + 1 = 0 implies costheta + 5sintheta = 1$$-\cos\theta - 5\sin\theta + 1 = 0 \implies \cos\theta + 5\sin\theta = 1$$ Let's express in half angles: 1 - 2sin^2fractheta2 + 10sinfractheta2cosfractheta2 = 1$$1 - 2\sin^2\frac{\theta}{2} + 10\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 1$$ 2sinfractheta2 left( 5cosfractheta2 - sinfractheta2 right) = 0$$2\sin\frac{\theta}{2} \left( 5\cos\frac{\theta}{2} - \sin\frac{\theta}{2} \right) = 0$$ So, sinfractheta2 = 0 implies costheta = 1$\sin\frac{\theta}{2} = 0 \implies \cos\theta = 1$ or tanfractheta2 = 5 implies costheta = frac1 - tan^2(theta/2)1 + tan^2(theta/2) = frac1 - 251 + 25 = -frac2426 = -frac1213$\tan\frac{\theta}{2} = 5 \implies \cos\theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - 25}{1 + 25} = -\frac{24}{26} = -\frac{12}{13}$. ### Step 4: Final Calculation The abscissa of P$P$ is 2costheta$2\cos\theta$. Values of abscissa are 2(1) = 2$2(1) = 2$ and 2left(-frac1213right) = -frac2413$2\left(-\frac{12}{13}\right) = -\frac{24}{13}$. Sum of abscissa values = 2 - frac2413 = frac26 - 2413 = frac213$= 2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$. We need 13 times (textSum) = 13 times frac213 = 2$13 \times (\text{Sum}) = 13 \times \frac{2}{13} = 2$. ### Pattern Recognition Instead of finding the image of a generic circle point in a line, construct the reflection point Q$Q$ parameter, use slope logic (m_1m_2=-1$m_1m_2=-1$) and midpoint logic simultaneously to create a trigonometric linear equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Straight Lines

### Related Formula Perpendicular distance from point (x_0, y_0)$(x_0, y_0)$ to line Ax + By + C = 0$Ax + By + C = 0$: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ ### Core Logic Since the circle touches two symmetric intersecting lines, its center must lie on their angle bisector (x$x$-axis). Use this property to find the center parameters.

### Step 1: Establish Center and Radius Equations The lines are x+y-3=0$x+y-3=0$ and x-y-3=0$x-y-3=0$. The intersection point is (3,0)$(3,0)$, and the bisector line is the x$x$-axis. Let the center be C(a, 0)$C(a, 0)$. The radius r$r$ is the perpendicular distance to either line: r = frac|a - 0 - 3|sqrt1^2 + 1^2 = frac|a - 3|sqrt2$$r = \frac{|a - 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{|a - 3|}{\sqrt{2}}$$ ### Step 2: Apply Point Passage Constraint The circle equation is (x - a)^2 + y^2 = r^2$(x - a)^2 + y^2 = r^2$. Substitute the given passage point (-9, 4)$(-9, 4)$: (-9 - a)^2 + 4^2 = left(fraca - 3sqrt2right)^2$$(-9 - a)^2 + 4^2 = \left(\frac{a - 3}{\sqrt{2}}\right)^2$$ 2left(a^2 + 18a + 81 + 16right) = a^2 - 6a + 9$$2\left(a^2 + 18a + 81 + 16\right) = a^2 - 6a + 9$$ 2a^2 + 36a + 194 = a^2 - 6a + 9 implies a^2 + 42a + 185 = 0$$2a^2 + 36a + 194 = a^2 - 6a + 9 \implies a^2 + 42a + 185 = 0$$ ### Step 3: Solve for Quadratic Roots Factor the quadratic equation: (a + 37)(a + 5) = 0 implies a_1 = -37, quad a_2 = -5$$(a + 37)(a + 5) = 0 \implies a_1 = -37, \quad a_2 = -5$$ ### Step 4: Compute Radii Squares Difference Find the corresponding radius value for each root: r_1 = frac|-37 - 3|sqrt2 = frac40sqrt2 = 20sqrt2 implies r_1^2 = 800$$r_1 = \frac{|-37 - 3|}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \implies r_1^2 = 800$$ r_2 = frac|-5 - 3|sqrt2 = frac8sqrt2 = 4sqrt2 implies r_2^2 = 32$$r_2 = \frac{|-5 - 3|}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \implies r_2^2 = 32$$ The absolute difference between their squares is: left|r_1^2 - r_2^2right| = |800 - 32| = 768$$\left|r_1^2 - r_2^2\right| = |800 - 32| = 768$$ ### Pattern Recognition Recognizing that the center must lie on the line of symmetry (x$x$-axis) eliminates one variable parameter immediately, reducing a difficult geometric system to a simple single-variable quadratic equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

### 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