Question 14 of 70beginner🔧 ApplyShort Answer2 marks

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Correct Answer

(x−2)2+(y−2)2=13(x - 2)^2 + (y - 2)^2 = 13

Exercise: EXERCISE 10.1 | Q: 14 | (Page 6)
For More Understanding

Explanation

The problem requires finding the equation of a circle given its centre and a point on the circumference. The textbook context provides the standard equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 where (h,k)(h, k) is the centre. It also establishes that the radius is the distance from the centre to any point on the circle. Following the method in Example 2, we substitute the centre coordinates. The radius is determined using the distance formula logic presented in the derivation section.

Solution Steps

  1. Step 1: Identify the centre (h,k)=(2,2)(h, k) = (2, 2) from the question.

  2. Step 2: Calculate the radius rr using the distance between the centre (2,2)(2, 2) and the point (4,5)(4, 5). r=(4−2)2+(5−2)2=22+32=4+9=13r = \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}.

  3. Step 3: Substitute h=2h=2, k=2k=2, and r2=13r^2 = 13 into the standard equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.

  4. Step 4: The final equation is (x−2)2+(y−2)2=13(x - 2)^2 + (y - 2)^2 = 13.

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