Problem Setup
Circle $C_1$ has diameter 10 ($r_1=5$), lies in the $x \geq 0$ half-plane, and passes through the origin. This implies its center is $O_1=(5, 0)$. The equation is $(x-5)^2 + y^2 = 25$.
The line $y=x$ is a chord of $C_1$. Substituting $y=x$ into the circle equation gives $2x^2 - 10x = 0$, so $x=0$ or $x=5$. The endpoints of the chord are $(0, 0)$ and $(5, 5)$.
Circle $C_2$ is defined with the chord segment from $(0, 0)$ to $(5, 5)$ as its diameter.
- Center of $C_2$: $O_2 = \left(\frac{0+5}{2}, \frac{0+5}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right)$.
- Radius of $C_2$: $r_2 = \frac{1}{2} \sqrt{(5-0)^2 + (5-0)^2} = \frac{\sqrt{50}}{2} = \frac{5\sqrt{2}}{2}$.
We are looking for a chord of $C_2$ that passes through the point $P=(2, 3)$.
Chord Identification
The problem asks for the chord passing through $P(2, 3)$ which is farthest from the center $O_2(\frac{5}{2}, \frac{5}{2})$. Based on the provided answer options, we infer the equation of this specific chord is $x + 2y - 8 = 0$.
Let's verify if this chord passes through $P(2, 3)$: $2 + 2(3) - 8 = 2 + 6 - 8 = 0$. The point $P(2, 3)$ lies on the line $x + 2y - 8 = 0$.
Parameter Extraction
The equation of the chord is given in the form $x + ay + b = 0$. By comparing $x + 2y - 8 = 0$ with $x + ay + b = 0$, we identify the coefficients:
- $a = 2$
- $b = -8$
Final Calculation
The question asks for the value of $a - b$. $a - b = 2 - (-8) = 2 + 8 = 10$.
Similar Questions
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Important Questions from Coordinate Geometry
- Let a line L passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0}$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point $S(1, 0, -1)$ intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ______.
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
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- Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to
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