Question 12 of 70intermediate🔧 ApplyLong Answer4 marks

Find the equation of the circle with radius 55 whose centre lies on xx-axis and passes through the point (2,3)(2, 3).

Correct Answer

Let the centre of the circle be (h,k)(h, k). Since the centre lies on the xx-axis, its yy-coordinate is 00, i.e., k=0k = 0. So the centre is (h,0)(h, 0). The radius is given as 55. The circle passes through the point (2,3)(2, 3). Therefore, the distance between the centre and this point equals the radius.

Using the distance formula, we have:

(2−h)2+(3−0)2=5\sqrt{(2 - h)^2 + (3 - 0)^2} = 5

Squaring both sides:

(2−h)2+9=25(2 - h)^2 + 9 = 25

Simplifying:

(2−h)2=16(2 - h)^2 = 16

Taking square root:

2−h=±42 - h = \pm 4

Hence, h=−2h = -2 or h=6h = 6.

Thus, there are two possible circles:

  1. Centre (−2,0)(-2, 0), radius 55: Equation is (x+2)2+y2=25(x + 2)^2 + y^2 = 25.
  2. Centre (6,0)(6, 0), radius 55: Equation is (x−6)2+y2=25(x - 6)^2 + y^2 = 25.

Both equations satisfy the given conditions.

Exercise: EXERCISE 10.1 | Q: 12 | (Page 6)
For More Understanding

Explanation

This question is from Exercise 10.1, Q12. The solution uses the standard equation of a circle (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, with centre (h,k)(h, k) and radius rr. Since the centre lies on the xx-axis, k=0k = 0. Substituting the given point and radius yields a quadratic in hh, giving two possible centres. Both circles are valid because the point (2,3)(2,3) is at a distance of 55 from two points on the xx-axis.

Solution Steps

  1. Step 1: Let centre be (h,k)(h, k). Since centre lies on xx-axis, k=0k = 0.

  2. Step 2: Use the condition that the circle passes through (2,3)(2,3) and has radius 55:

    (2−h)2+(3−0)2=52(2 - h)^2 + (3 - 0)^2 = 5^2

  3. Step 3: Simplify:

    (2−h)2+9=25(2 - h)^2 + 9 = 25 (2−h)2=16(2 - h)^2 = 16 2−h=±42 - h = \pm 4 h=−2 or h=6h = -2 \text{ or } h = 6

  4. Step 4: Write the equations: For h=−2h = -2,

    (x+2)2+y2=25(x + 2)^2 + y^2 = 25

    for h=6h = 6,

    (x−6)2+y2=25(x - 6)^2 + y^2 = 25
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