Equation of two diameters of a circle are 2x-3y=5$2x-3y=5$ and 3x-4y=7$3x-4y=7$. The line joining the points (-frac227,-4)$(-\frac{22}{7},-4)$ and (-frac17,3)$(-\frac{1}{7},3)$ intersects the circle at only one point P(alpha,beta)$P(\alpha,\beta)$. Then 17beta-alpha$17\beta-\alpha$ is equal to

### 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