Question

Among the chords of the circle $x^2+y^2=75$, the number of chords having their midpoints on the line $x=8$ and having their slopes as integers is

• A. 8
• B. 6
• C. 4
• D. 2

Hint:

For a circle centered at the origin, the slope of a chord with midpoint $(h,k)$ is $-h/k$. This comes from the fact that the radius to the midpoint is perpendicular to the chord. This is often simpler than using the $T=S_1$ formula.

Answer 1

The Correct Option is C

Step 1: Relate the midpoint of a chord to its slope.
Let the midpoint of a chord be $M(h,k)$. The circle is $x^2+y^2=75$, centered at the origin $O(0,0)$. The line segment OM connecting the center to the midpoint is perpendicular to the chord. Slope of OM is $m_{OM} = \frac{k-0}{h-0} = \frac{k}{h}$. The slope of the chord, $m$, is the negative reciprocal of the slope of OM. \[ m = -\frac{1}{m_{OM}} = -\frac{h}{k}. \] 

Step 2: Apply the given conditions. 
The midpoint $M(h,k)$ lies on the line $x=8$, so $h=8$. The slope of the chord is $m = -\frac{8}{k}$. We are given that $m$ must be an integer. This implies that $k$ must be an integer divisor of 8. The possible values for $k$ are $\{\pm 1, \pm 2, \pm 4, \pm 8\}$. 

Step 3: Apply the condition that the midpoint must be inside the circle. 
For a chord to exist, its midpoint must be strictly inside the circle. The condition is $h^2+k^2<R^2$. Given $h=8$ and $R^2=75$: \[ 8^2 + k^2<75. \] \[ 64 + k^2<75. \] \[ k^2<11. \] 

Step 4: Find the valid values of k and count the chords. 
We need to find the integer values of $k$ from the set $\{\pm 1, \pm 2, \pm 4, \pm 8\}$ that also satisfy $k^2<11$. If $k = \pm 1$, then $k^2 = 1<11$. These are valid. (2 values) If $k = \pm 2$, then $k^2 = 4<11$. These are valid. (2 values) If $k = \pm 3$, $k^2=9<11$, but 3 is not a divisor of 8. If $k = \pm 4$, then $k^2 = 16>11$. Not valid. If $k = \pm 8$, then $k^2 = 64>11$. Not valid. The valid integer values for $k$ are $-2, -1, 1, 2$. Each distinct midpoint $(8,k)$ defines a unique chord. Therefore, there are 4 such chords.