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Question: There are two circles passing through points A(-1, 2) and B(2, 3) having radius 5 . Then the length ...

There are two circles passing through points A(-1, 2) and B(2, 3) having radius 5 . Then the length of intercept on x-axis of the circle intersecting x-axis is :

A

2√21

Answer

2√21

Explanation

Solution

The problem asks us to find the length of the x-intercept of a circle that passes through two given points A and B and has a specified radius.

1. Find the equations of the circles:

Let the center of the circle be (h, k) and the radius be R = 5. The equation of the circle is (x−h)2+(y−k)2=R2(x-h)^2 + (y-k)^2 = R^2.
Since the circle passes through A(-1, 2) and B(2, 3), we can substitute these points into the equation:

For point A(-1, 2):
(−1−h)2+(2−k)2=52(-1-h)^2 + (2-k)^2 = 5^2
(1+h)2+(2−k)2=25(1+h)^2 + (2-k)^2 = 25
1+2h+h2+4−4k+k2=251 + 2h + h^2 + 4 - 4k + k^2 = 25
h2+k2+2h−4k+5=25h^2 + k^2 + 2h - 4k + 5 = 25
h2+k2+2h−4k−20=0h^2 + k^2 + 2h - 4k - 20 = 0 (Equation 1)

For point B(2, 3):
(2−h)2+(3−k)2=52(2-h)^2 + (3-k)^2 = 5^2
4−4h+h2+9−6k+k2=254 - 4h + h^2 + 9 - 6k + k^2 = 25
h2+k2−4h−6k+13=25h^2 + k^2 - 4h - 6k + 13 = 25
h2+k2−4h−6k−12=0h^2 + k^2 - 4h - 6k - 12 = 0 (Equation 2)

2. Solve for h and k:

Subtract Equation 2 from Equation 1:
(h2+k2+2h−4k−20)−(h2+k2−4h−6k−12)=0(h^2 + k^2 + 2h - 4k - 20) - (h^2 + k^2 - 4h - 6k - 12) = 0
2h−4k−20+4h+6k+12=02h - 4k - 20 + 4h + 6k + 12 = 0
6h+2k−8=06h + 2k - 8 = 0
3h+k−4=03h + k - 4 = 0
From this, we get k=4−3hk = 4 - 3h.

Substitute this value of k into Equation 1:
h2+(4−3h)2+2h−4(4−3h)−20=0h^2 + (4 - 3h)^2 + 2h - 4(4 - 3h) - 20 = 0
h2+(16−24h+9h2)+2h−16+12h−20=0h^2 + (16 - 24h + 9h^2) + 2h - 16 + 12h - 20 = 0
10h2+(−24h+2h+12h)+(16−16−20)=010h^2 + (-24h + 2h + 12h) + (16 - 16 - 20) = 0
10h2−10h−20=010h^2 - 10h - 20 = 0
Divide by 10:
h2−h−2=0h^2 - h - 2 = 0
Factor the quadratic equation:
(h−2)(h+1)=0(h - 2)(h + 1) = 0
This gives two possible values for h: h=2h = 2 or h=−1h = -1.

Find the corresponding k values:

  • If h=2h = 2, then k=4−3(2)=4−6=−2k = 4 - 3(2) = 4 - 6 = -2.
    Center C1 = (2, -2).
  • If h=−1h = -1, then k=4−3(−1)=4+3=7k = 4 - 3(-1) = 4 + 3 = 7.
    Center C2 = (-1, 7).

So, the two circles are:
Circle 1: (x−2)2+(y+2)2=52=25(x-2)^2 + (y+2)^2 = 5^2 = 25
Circle 2: (x+1)2+(y−7)2=52=25(x+1)^2 + (y-7)^2 = 5^2 = 25

3. Determine which circle intersects the x-axis:

A circle intersects the x-axis if the absolute value of its y-coordinate of the center (∣k∣|k|) is less than or equal to its radius (R).

  • For Circle 1 (Center (2, -2), R=5):
    ∣k∣=∣−2∣=2|k| = |-2| = 2. Since 2<52 < 5, Circle 1 intersects the x-axis.

  • For Circle 2 (Center (-1, 7), R=5):
    ∣k∣=∣7∣=7|k| = |7| = 7. Since 7>57 > 5, Circle 2 does not intersect the x-axis.

Therefore, only Circle 1 intersects the x-axis.

4. Calculate the length of the x-intercept for Circle 1:

The equation of Circle 1 is (x−2)2+(y+2)2=25(x-2)^2 + (y+2)^2 = 25.
To find the x-intercept, set y = 0:
(x−2)2+(0+2)2=25(x-2)^2 + (0+2)^2 = 25
(x−2)2+4=25(x-2)^2 + 4 = 25
(x−2)2=21(x-2)^2 = 21
x−2=±21x-2 = \pm\sqrt{21}
x=2±21x = 2 \pm\sqrt{21}

The x-intercepts are x1=2−21x_1 = 2 - \sqrt{21} and x2=2+21x_2 = 2 + \sqrt{21}.
The length of the intercept is the distance between these two points:
Length =∣x2−x1∣=∣(2+21)−(2−21)∣= |x_2 - x_1| = |(2 + \sqrt{21}) - (2 - \sqrt{21})|
Length =∣2+21−2+21∣= |2 + \sqrt{21} - 2 + \sqrt{21}|
Length =∣221∣= |2\sqrt{21}|
Length =221= 2\sqrt{21}

Alternatively, for a circle (x−h)2+(y−k)2=R2(x-h)^2 + (y-k)^2 = R^2, the length of the x-intercept is given by the formula 2R2−k22\sqrt{R^2 - k^2}.
For Circle 1, h=2, k=-2, R=5.
Length =252−(−2)2= 2\sqrt{5^2 - (-2)^2}
Length =225−4= 2\sqrt{25 - 4}
Length =221= 2\sqrt{21}

The final answer is 221\boxed{2\sqrt{21}}.