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Exercise 10.1 · Q14

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

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Use the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 with centre (h,k)=(2,2)(h, k) = (2, 2); find the radius from the distance to (4,5)(4, 5), giving (x−2)2+(y−2)2=13(x - 2)^2 + (y - 2)^2 = 13.

The standard equation of a circle is built around its defining property: every point on the circle is exactly the same distance (the radius) from the centre. If the centre is at (h,k)(h, k) and the radius is rr, then any point (x,y)(x, y) on the circle satisfies

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

This is just the distance formula squared. We know the centre, so we need only find r2r^2 by using the fact that the given point (4,5)(4, 5) lies on the circle.

Step-by-step construction:

  1. Identify the centre coordinates. We have h=2h = 2 and k=2k = 2, so the equation takes the form

(x−2)2+(y−2)2=r2.(x - 2)^2 + (y - 2)^2 = r^2.

  1. Calculate the radius using the point (4,5)(4, 5). The distance from the centre (2,2)(2, 2) to the point (4,5)(4, 5) is the radius. Apply the distance formula:

r=(4−2)2+(5−2)2=22+32=4+9=13.r = \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}.

  1. Square the radius to get r2r^2. Since the standard form uses r2r^2, we have

r2=13.r^2 = 13.

  1. Write the final equation. Substitute r2=13r^2 = 13 into the equation from step 1: …

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