If a circle passing through the point $(-1, 0)$ touches y-axis at $(0, 2)$, then the length of the chord of the circle along the x-axis is
- $\displaystyle \frac{5}{2}$
- $5$
- $\displaystyle \frac{3}{2}$
- $3$
The circle touches the y-axis at (0, 2), so its center is (r, 2). Since it passes through (-1, 0), the distance from (r, 2) to (-1, 0) equals r. Solving (r+1)^2 + (2-0)^2 = r^2 gives r = -2.5. The circle equation is (x+2.5)^2 + (y-2)^2 = 6.25. Setting y=0, (x+2.5)^2 = 6.25 - 4 = 2.25, so x+2.5 = +/- 1.5. The chord length is |(-1) - (-4)| = 3.
A circle touching the y-axis at (0, 2) has its center at a horizontal distance equal to its radius from the y-axis, so if the center is (h, k), then |h| = r and k = 2. The center to the right of the y-axis is at (r, 2) because the circle passes through (-1, 0), which is on the left side of the y-axis, meaning the center must be at (-r, 2). The distance from the center (-r, 2) to the point of tangency (0, 2) gives the radius as | -r - 0 | = r. We use the fact that the circle passes through (-1, 0) to set up the distance formula: sqrt((-1 - (-r))^2 + (0 - 2)^2) = r. Squaring both sides gives (-1 + r)^2 + (-2)^2 = r^2, which expands to 1 - 2r + r^2 + 4 = r^2. Solving for r, the r^2 terms cancel out, leaving 5 - 2r = 0, so r = 5/2. The center of the circle is (-5/2, 2) and the radius is 5/2. The length of the chord along the x-axis is found by setting y = 0 in the circle's equation. The horizontal distance from the center to the x-axis chord is the y-coordinate of the center, which is 2. Using the half-chord length formula, L/2 = sqrt(r^2 - (distance)^2) = sqrt((5/2)^2 - 2^2) = sqrt(25/4 - 16/4) = sqrt(9/4) = 3/2. The total length of the chord is twice this value, which is 3. The result is 3.