The absolute difference between the squares of the radii of the two circles passing through the point (-9, 4)$(-9, 4)$ and touching the lines x + y = 3$x + y = 3$ and x - y = 3$x - y = 3$, is equal to ________.

### Related Formula The coordinates for the reflection image of a point (x_1, y_1)$(x_1, y_1)$ across a standard line ax + by + c = 0$ax + by + c = 0$ are determined using: fracx - x_1a = fracy - y_1b = frac-2(ax_1 + by_1 + c)a^2 + b^2$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$ ### Core Logic Find the center and radius of the original given circle: x^2 + y^2 - 2x + 4y - 4 = 0 implies textCenter = (1, -2), \, r = sqrt1^2 + (-2)^2 - (-4) = 3$$x^2 + y^2 - 2x + 4y - 4 = 0 \implies \text{Center} = (1, -2), \, r = \sqrt{1^2 + (-2)^2 - (-4)} = 3$$

Reflect the center point (1, -2)$(1, -2)$ across the line mirror 2x - 3y + 5 = 0$2x - 3y + 5 = 0$: fracx - 12 = fracy + 2-3 = frac-2(2(1) - 3(-2) + 5)2^2 + (-3)^2 = frac-2(2 + 6 + 5)13 = -2$$\frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{-2(2(1) - 3(-2) + 5)}{2^2 + (-3)^2} = \frac{-2(2 + 6 + 5)}{13} = -2$$ x - 1 = -4 implies x = -3$$x - 1 = -4 \implies x = -3$$ y + 2 = 6 implies y = 4$$y + 2 = 6 \implies y = 4$$ Thus, the center O$O$ of the reflected circle C$C$ is (-3, 4)$(-3, 4)$, and its radius is preserved at r = 3$r = 3$. ### Step 1: Locate Point A We are given that OA$OA$ is \parallel to the x-axis, meaning its y-coordinate matches the center. Since A$A$ lies to the \right of the center O(-3, 4)$O(-3, 4)$: A = (-3 + r, \, 4) = (-3 + 3, \, 4) = (0, 4)$$A = (-3 + r, \, 4) = (-3 + 3, \, 4) = (0, 4)$$ ### Step 2: Determine Angular Position of Point B The arc length AB$AB$ is given as frac16$\frac{1}{6}$ of the total perimeter: textArc length = rtheta = frac16(2pi r) implies theta = fracpi3 = 60^circ$$\text{Arc length} = r\theta = \frac{1}{6}(2\pi r) \implies \theta = \frac{\pi}{3} = 60^\circ$$

Using parametric coordinates relative to center O(-3, 4)$O(-3, 4)$ with radius r = 3$r = 3$: alpha = -3 + 3costheta, quad beta = 4 + 3sintheta$$\alpha = -3 + 3\cos\theta, \quad \beta = 4 + 3\sin\theta$$ Since beta < 4$\beta < 4$, the \angle theta$\theta$ must point downwards into the negative quadrant relative to A$A$, meaning theta = -60^circ = -fracpi3$\theta = -60^\circ = -\frac{\pi}{3}$: alpha = -3 + 3cosleft(-fracpi3right) = -3 + 3left(frac12 ight) = -frac32$$\alpha = -3 + 3\cos\left(-\frac{\pi}{3}\right) = -3 + 3\left(\frac{1}{2} ight) = -\frac{3}{2}$$ beta = 4 + 3sinleft(-fracpi3right) = 4 - frac3sqrt32$$\beta = 4 + 3\sin\left(-\frac{\pi}{3}\right) = 4 - \frac{3\sqrt{3}}{2}$$ ### Step 3: Evaluate Final Algebraic Value Substitute the determined coordinates into the target expression: beta - sqrt3alpha = left(4 - frac3sqrt32right) - sqrt3left(-frac32 ight)$$\beta - \sqrt{3}\alpha = \left(4 - \frac{3\sqrt{3}}{2}\right) - \sqrt{3}\left(-\frac{3}{2} ight)$$ beta - sqrt3alpha = 4 - frac3sqrt32 + frac3sqrt32 = 4$$\beta - \sqrt{3}\alpha = 4 - \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 4$$ ### Pattern Recognition Whenever parametric configurations on a circle involve radical coordinate multipliers like beta - sqrt3alpha$\beta - \sqrt{3}\alpha$, using angular vectors centered at the origin of the circle avoids setting up and solving long distance equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles

### Related Formula textArea of a quadrilateral with perpendicular diagonals d_1 text and d_2 = frac12 d_1 d_2$$\text{Area of a quadrilateral with perpendicular diagonals } d_1 \text{ and } d_2 = \frac{1}{2} d_1 d_2$$ ### Core Logic The line perpendicular to chord AB$AB$ passing through its midpoint is the diameter of the circle because the perpendicular bisector of any chord passes through the center. Thus, CD$CD$ is a diameter, making its length equal to 2R = 4$2R = 4$.

### Step 1: Find Length of Chord AB The perpendicular distance p$p$ from the center (0,0)$(0,0)$ to the line x + y - 1 = 0$x + y - 1 = 0$ is: p = frac|0 + 0 - 1|sqrt1^2 + 1^2 = frac1sqrt2$$p = \frac{|0 + 0 - 1|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$$ Length of chord AB = 2sqrtR^2 - p^2 = 2sqrt4 - frac12 = 2sqrtfrac72 = sqrt14$AB = 2\sqrt{R^2 - p^2} = 2\sqrt{4 - \frac{1}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{14}$. ### Step 2: Area Calculation Since CD$CD$ is the perpendicular bisector of AB$AB$, the diagonals of quadrilateral ADBC$ADBC$ are perpendicular. Therefore, the area is: textArea = frac12 times AB times CD = frac12 times sqrt14 times 4 = 2sqrt14$$\text{Area} = \frac{1}{2} \times AB \times CD = \frac{1}{2} \times \sqrt{14} \times 4 = 2\sqrt{14}$$ ### Pattern Recognition Whenever a line passes through the midpoint of a chord and is perpendicular to it, recognize instantly that it is a diameter. The area of the quadrilateral formed is then simply the area of two triangles sharing the diameter as a base, or frac12 d_1 d_2$\frac{1}{2} d_1 d_2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

### Related Formula (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$ This is the equation of a circle passing through diametrically opposite endpoints (x_1, y_1)$(x_1, y_1)$ and (x_2, y_2)$(x_2, y_2)$. ### Core Logic Look at the three known points: A(1, 0)$A(1, 0)$, B(0, 1)$B(0, 1)$, and O(0, 0)$O(0, 0)$. The vectors vecOA$\vec{OA}$ and vecOB$\vec{OB}$ align with the x and y axes, meaning angle AOB = 90^circ$\angle AOB = 90^{\circ}$. Since the angle subtended by AB$AB$ at a point O$O$ on the circle is 90^circ$90^{\circ}$, the segment AB$AB$ must be the diameter of the circle. ### Step 1: Finding the Circle Equation Using the diametric form for points A(1,0)$A(1,0)$ and B(0,1)$B(0,1)$: (x-1)(x-0) + (y-0)(y-1) = 0$$(x-1)(x-0) + (y-0)(y-1) = 0$$ x^2 - x + y^2 - y = 0$$x^2 - x + y^2 - y = 0$$ ### Step 2: Solving for k Since the point (2k, 3k)$(2k, 3k)$ also lies on this circle, substitute x = 2k$x = 2k$ and y = 3k$y = 3k$ into the equation: (2k)^2 - (2k) + (3k)^2 - (3k) = 0$$(2k)^2 - (2k) + (3k)^2 - (3k) = 0$$ 4k^2 - 2k + 9k^2 - 3k = 0$$4k^2 - 2k + 9k^2 - 3k = 0$$ 13k^2 - 5k = 0$13k^2 - 5k = 0$ k(13k - 5) = 0$k(13k - 5) = 0$ ### Step 3: Finding valid points This gives two possible values for k$k$: k = 0$k = 0$ or k = frac513$k = \frac{5}{13}$. If k=0$k=0$, the fourth point becomes (0,0)$(0,0)$, which is already listed. The question specifies four *distinct* points. Thus, k = frac513$k = \frac{5}{13}$. ### Pattern Recognition Whenever (0,0)$(0,0)$, (a,0)$(a,0)$, and (0,b)$(0,b)$ lie on a circle, they form a right triangle at the origin. The hypotenuse connecting (a,0)$(a,0)$ and (0,b)$(0,b)$ is ALWAYS the diameter. You can immediately write the circle's equation as x^2 + y^2 - ax - by = 0$x^2 + y^2 - ax - by = 0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles

### Related Formula The intersection of any two non-parallel diameters yields the center of the circle. A line that intersects a circle at exactly one point is a tangent line. The tangent at the point of contact P$P$ is always perpendicular to the radius CP$CP$, thus m_texttangent times m_textradius = -1$m_{\text{tangent}} \times m_{\text{radius}} = -1$. ### Core Logic Find the center C$C$ by solving the two diameter equations: 2x - 3y = 5 quad dots (1)$$2x - 3y = 5 \quad \dots (1)$$ 3x - 4y = 7 quad dots (2)$$3x - 4y = 7 \quad \dots (2)$$ Multiply (1) by 3 and (2) by 2: 6x - 9y = 15$6x - 9y = 15$ 6x - 8y = 14$6x - 8y = 14$ Subtracting the equations gives -y = 1 Rightarrow y = -1$-y = 1 \Rightarrow y = -1$. Substitute y = -1$y = -1$ into (1): 2x + 3 = 5 Rightarrow 2x = 2 Rightarrow x = 1$2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1$. Center C$C$ is (1, -1)$(1, -1)$. Find the equation of the line joining A(-frac227, -4)$A(-\frac{22}{7}, -4)$ and B(-frac17, 3)$B(-\frac{1}{7}, 3)$. Slope of AB$AB$: m_AB = frac3 - (-4)-1/7 - (-22/7) = frac721/7 = frac73$m_{AB} = \frac{3 - (-4)}{-1/7 - (-22/7)} = \frac{7}{21/7} = \frac{7}{3}$. Equation of line AB$AB$: y - 3 = frac73left(x + frac17right)$$y - 3 = \frac{7}{3}\left(x + \frac{1}{7}\right)$$ 3y - 9 = 7x + 1$3y - 9 = 7x + 1$ 7x - 3y + 10 = 0 quad dots text(Line AB)$$7x - 3y + 10 = 0 \quad \dots \text{(Line AB)}$$

### Step 1: Exploit Tangency Geometry Since line AB$AB$ intersects the circle at only one point P(alpha, beta)$P(\alpha, \beta)$, line AB$AB$ is a tangent to the circle, and P$P$ is the point of tangency. The radius line CP$CP$ is perpendicular to tangent AB$AB$. Slope of CP$CP$ (m_CP$m_{CP}$) must be -frac37$-\frac{3}{7}$. Equation of the line passing through center C(1, -1)$C(1, -1)$ with slope -frac37$-\frac{3}{7}$: y - (-1) = -frac37(x - 1)$$y - (-1) = -\frac{3}{7}(x - 1)$$ 7y + 7 = -3x + 3$$7y + 7 = -3x + 3$$ 3x + 7y + 4 = 0 quad dots text(Line CP)$$3x + 7y + 4 = 0 \quad \dots \text{(Line CP)}$$ ### Step 2: Solve for Intersection P Point P(alpha, beta)$P(\alpha, \beta)$ is the intersection of Tangent AB$AB$ and Radius CP$CP$. Solve the system: 7x - 3y = -10 quad dots (times 7)$$7x - 3y = -10 \quad \dots (\times 7)$$ 3x + 7y = -4 quad dots (times 3)$$3x + 7y = -4 \quad \dots (\times 3)$$ 49x - 21y = -70$49x - 21y = -70$ 9x + 21y = -12$9x + 21y = -12$ Add them: 58x = -82 Rightarrow x = -frac8258 = -frac4129$58x = -82 \Rightarrow x = -\frac{82}{58} = -\frac{41}{29}$. So, alpha = -frac4129$\alpha = -\frac{41}{29}$. Substitute x$x$ into 3x + 7y = -4$3x + 7y = -4$: 3left(-frac4129right) + 7y = -4$$3\left(-\frac{41}{29}\right) + 7y = -4$$ -frac12329 + 7y = -frac11629$$-\frac{123}{29} + 7y = -\frac{116}{29}$$ 7y = frac123 - 11629 = frac729 Rightarrow y = frac129$. So, $beta = frac129$. ### Step 3: Evaluate Target Expression Evaluate $17beta - alpha$: $$7y = \frac{123 - 116}{29} = \frac{7}{29} \Rightarrow y = \frac{1}{29}$. So, $\beta = \frac{1}{29}$. ### Step 3: Evaluate Target Expression Evaluate $17\beta - \alpha$: $$17\left(\frac{1}{29}\right) - \left(-\frac{41}{29}\right) = \frac{17 + 41}{29} = \frac{58}{29} = 2$$ ### Pattern Recognition When a line "intersects a circle at exactly one point", it's a coded cue to stop thinking about quadratics and discriminants, and immediately build a perpendicular geometric radius from the center to find the exact tangency coordinate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

### Related Formula textArea of triangle OAP: Delta = frac12 a b sin theta$$\text{Area of triangle } OAP: \Delta = \frac{1}{2} a b \sin \theta$$ textCommon Chord geometry: Height of triangle acts as half the common chord, so Delta = frac12 times textbase times left(fracbeta2right)$$\text{Common Chord geometry: Height of triangle acts as half the common chord, so } \Delta = \frac{1}{2} \times \text{base} \times \left(\frac{\beta}{2}\right)$$ ### Core Logic Radius of C_1$C_1$ is r_1 = 5$r_1 = 5$ (centered at Origin O(0,0)$O(0,0)$). Radius of C_2$C_2$ is r_2 = 4$r_2 = 4$ (centered at A(alpha,0)$A(\alpha,0)$ with 5 lt alpha lt 9$5 \lt \alpha \lt 9$). Let P$P$ be an intersection point. In Delta OAP$\Delta OAP$, the sides are OP = 5$OP = 5$, AP = 4$AP = 4$, and OA = alpha$OA = \alpha$. The angle between the radii at P$P$ is angle OPA = theta$\angle OPA = \theta$, where sin theta = fracsqrt638$\sin \theta = \frac{\sqrt{63}}{8}$. ### Step 1: Determining Area via the Sine Rule The area of Delta OAP$\Delta OAP$ can be computed using the two radii and the angle between them: textArea = frac12 (OP) (AP) sin theta$$\text{Area} = \frac{1}{2} (OP) (AP) \sin \theta$$ textArea = frac12 (5) (4) left( fracsqrt638 right) = 10 left( fracsqrt638 right) = frac5sqrt634$$\text{Area} = \frac{1}{2} (5) (4) \left( \frac{\sqrt{63}}{8} \right) = 10 \left( \frac{\sqrt{63}}{8} \right) = \frac{5\sqrt{63}}{4}$$ ### Step 2: Linking Area to the Common Chord

The base of Delta OAP$\Delta OAP$ on the x-axis is OA = alpha$OA = \alpha$. The height of Delta OAP$\Delta OAP$ dropped from P$P$ to OA$OA$ is exactly half the length of the common chord, which is fracbeta2$\frac{\beta}{2}$. textArea = frac12 times textbase times textheight = frac12 alpha left( fracbeta2 right) = fracalphabeta4$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \alpha \left( \frac{\beta}{2} \right) = \frac{\alpha\beta}{4}$$ ### Step 3: Equating Areas Equate the two area expressions: fracalphabeta4 = frac5sqrt634$$\frac{\alpha\beta}{4} = \frac{5\sqrt{63}}{4}$$ alphabeta = 5sqrt63$$\alpha\beta = 5\sqrt{63}$$ ### Step 4: Final Evaluation Square the result as requested: (alphabeta)^2 = (5sqrt63)^2 = 25 times 63 = 1575$$(\alpha\beta)^2 = (5\sqrt{63})^2 = 25 \times 63 = 1575$$ ### Pattern Recognition In intersecting circle problems, the triangle formed by the centers and the intersection point handles both the intersection angle (via cosine/sine area rules) and the common chord (which serves as a perpendicular height doubled). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Properties of Triangles