Question 2 of 70beginner🔧 ApplyShort Answer1 mark

In each of the following Exercises 1 to 5, find the equation of the circle with centre (-2, 3) and radius 4

Correct Answer

(x+2)2+(y−3)2=16(x + 2)^2 + (y - 3)^2 = 16

Exercise: EXERCISE 10.1 | Q: 2 | (Page 6)
For More Understanding

Explanation

Using the standard equation of a circle (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 where (h, k) is the centre and r is the radius. Given centre (-2, 3) means h = -2, k = 3, and radius = 4. Substituting: (x−(−2))2+(y−3)2=42(x - (-2))^2 + (y - 3)^2 = 4^2, which simplifies to (x+2)2+(y−3)2=16(x + 2)^2 + (y - 3)^2 = 16.

Solution Steps

  1. Step 1: Identify centre (h, k) = (-2, 3) and radius r = 4

  2. Step 2: Apply the standard equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

  3. Step 3: Substitute values: (x−(−2))2+(y−3)2=42(x - (-2))^2 + (y - 3)^2 = 4^2

  4. Step 4: Simplify to get (x+2)2+(y−3)2=16(x + 2)^2 + (y - 3)^2 = 16

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