### 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 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

### Related Formula textIntercept form of line: fracxa + fracyb = 1$\text{Intercept form of line: } \frac{x}{a} + \frac{y}{b} = 1$ ### Core Logic Circle x^2 + y^2 - 16x - 4y = 0$x^2 + y^2 - 16x - 4y = 0$ has its centre at (8, 2)$(8, 2)$. Let the line passing through (8, 2)$(8, 2)$ have slope m$m$. Its equation is: y - 2 = m(x - 8)$y - 2 = m(x - 8)$ x-intercept (A$A$): set y=0 implies -2 = m(x-8) implies x = 8 - frac2m$y=0 \implies -2 = m(x-8) \implies x = 8 - \frac{2}{m}$. y-intercept (B$B$): set x=0 implies y = 2 - 8m$x=0 \implies y = 2 - 8m$. Sum of intercepts OA + OB = (8 - frac2m) + (2 - 8m) = 10 - frac2m - 8m$OA + OB = (8 - \frac{2}{m}) + (2 - 8m) = 10 - \frac{2}{m} - 8m$. To minimize, let f(m) = 10 - frac2m - 8m$f(m) = 10 - \frac{2}{m} - 8m$. f'(m) = frac2m^2 - 8 = 0 implies m^2 = frac14$f'(m) = \frac{2}{m^2} - 8 = 0 \implies m^2 = \frac{1}{4}$ Since the line meets the positive coordinate axes, intercepts must be positive, which requires m < 0$m < 0$. Thus m = -1/2$m = -1/2$. Substitute m = -1/2$m = -1/2$: OA + OB = 10 - frac2-1/2 - 8(-1/2) = 10 + 4 + 4 = 18$OA + OB = 10 - \frac{2}{-1/2} - 8(-1/2) = 10 + 4 + 4 = 18$ ### Pattern Recognition AM-GM can also be applied: 8a + 2b = ab implies 1 = frac8a + frac2b$8a + 2b = ab \implies 1 = \frac{8}{a} + \frac{2}{b}$. To minimize a+b$a+b$, use Cauchy-Schwarz or standard differentiation. Differentiation directly yields intercept minima. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles