Let $y=x$ be the equation of a chord of the circle $C_{1}$ (in the closed half-plane $x\ge0$) of diameter 10 passing through the origin. Let $C_{2}$ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $C_{2}$ which passes through the point $(2, 3)$ and is farthest from the center of $C_{2}$ is $x+ay+b=0$, then $a-b$ is equal to:
- (1) 10
- (2) -6
- (3) -2
- (4) 6
Solution:
First, find the equation of circle $C_1$. It has diameter 10 and passes through the origin. Since it lies in $x \ge 0$ and has the chord $y=x$ passing through origin, the circle is symmetric about some axis. Assuming standard orientation tangent to y-axis or passing through origin such that diameter is along x-axis (based on typical problem structures and diagram), the center is $(5,0)$ and radius is 5.
Equation of $C_1$: $(x-5)^2 + y^2 = 5^2 \Rightarrow x^2 + y^2 – 10x = 0$.
Find the intersection points of $C_1$ and the line $y=x$:
$$x^2 + x^2 – 10x = 0 \Rightarrow 2x^2 – 10x = 0$$
$$2x(x-5) = 0 \Rightarrow x=0, x=5$$
The points are $O(0,0)$ and $P(5,5)$.
$C_2$ has $OP$ as its diameter. The equation of $C_2$ is:
$$(x-0)(x-5) + (y-0)(y-5) = 0$$
$$x^2 – 5x + y^2 – 5y = 0$$
The center of $C_2$ is $N(\frac{5}{2}, \frac{5}{2})$.
We need the chord of $C_2$ passing through $B(2,3)$ that is farthest from the center $N$. This chord is perpendicular to the line segment $NB$.
Slope of $NB$:
$$m_{NB} = \frac{3 – 2.5}{2 – 2.5} = \frac{0.5}{-0.5} = -1$$
Since the required chord is perpendicular to $NB$, its slope $m$ is:
$$m = -\frac{1}{m_{NB}} = -\frac{1}{-1} = 1$$
Equation of the chord passing through $B(2,3)$ with slope 1:
$$y – 3 = 1(x – 2)$$
$$y – 3 = x – 2$$
$$x – y + 1 = 0$$
Comparing this with $x + ay + b = 0$:
$$a = -1, \quad b = 1$$
We need to find $a – b$:
$$a – b = -1 – 1 = -2$$
Ans. (3)