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Exercise 10.1 · Q1

Q.Find the equation of the circle with centre (0,2)(0, 2) and radius 22.

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✓ Free question

A circle is the set of all points at a fixed distance (radius) from a center; substituting center (0,2)(0, 2) and radius 22 into the standard form gives (x−0)2+(y−2)2=4(x - 0)^2 + (y - 2)^2 = 4.

The standard form of a circle's equation captures its geometric definition perfectly: every point (x,y)(x, y) on the circle is exactly rr units away from the center (h,k)(h, k). The distance formula (x−h)2+(y−k)2=r\sqrt{(x - h)^2 + (y - k)^2} = r squares to eliminate the root, giving us the clean algebraic form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.

This form makes sense because the left side measures the squared distance from any point (x,y)(x, y) to the center, and the right side is the squared radius — so the equation says "all points whose squared distance from the center equals the squared radius," which is precisely what a circle is.

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

where (h,k)(h, k) is the center and rr is the radius.

Now we substitute the given values directly.

  1. Identify the center coordinates: We have h=0h = 0 and k=2k = 2.

  2. Identify the radius: We have r=2r = 2, so r2=4r^2 = 4.

  3. Substitute into the standard form:

(x−0)2+(y−2)2=4(x - 0)^2 + (y - 2)^2 = 4

  1. Simplify: Since x−0=xx - 0 = x, we get

x2+(y−2)2=4x^2 + (y - 2)^2 = 4

You could expand (y−2)2(y - 2)^2 to write this as x2+y2−4y+4=4x^2 + y^2 - 4y + 4 = 4, which simplifies to x2+y2−4y=0x^2 + y^2 - 4y = 0, but the form x2+(y−2)2=4x^2 + (y - 2)^2 = 4 is cleaner and immediately reveals the circle's center and radius.

Tip

When the center has a zero coordinate, the corresponding term simplifies beautifully: (x−0)2=x2(x - 0)^2 = x^2. This circle sits on the yy-axis, centered at (0,2)(0, 2).

✓Final answer

The equation of the circle is x2+(y−2)2=4x^2 + (y - 2)^2 = 4.

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