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Exercise 10.1 · Q4

Q.Find the equation of the circle with centre (1,1)(1, 1) and radius 2\sqrt{2}.

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The standard form of a circle is (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2. Substituting centre (1,1)(1,1) and radius 2\sqrt{2} gives (x−1)2+(y−1)2=2(x-1)^2+(y-1)^2=2, which simplifies to x2+y2−2x−2y=0x^2+y^2-2x-2y=0.

The equation of a circle is simply a way of saying: all points (x,y)(x,y) that are exactly a fixed distance (the radius) from a fixed point (the centre). This is the geometric definition, and the algebra follows directly from the distance formula.

If the centre is (h,k)(h,k) and the radius is rr, then for any point (x,y)(x,y) on the circle, the distance between (x,y)(x,y) and (h,k)(h,k) must equal rr. The distance formula gives:

(x−h)2+(y−k)2=r\sqrt{(x-h)^2+(y-k)^2}=r

Squaring both sides removes the square root and gives the clean standard form:

(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2

(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2

This is the only formula you need. No memorising expansions — just plug in and simplify if asked.

Now let’s apply it to the given numbers.

  1. Identify the centre and radius.

    The centre is (h,k)=(1,1)(h,k)=(1,1) and the radius is r=2r=\sqrt{2}.

  2. Substitute into the standard form.

(x−1)2+(y−1)2=(2)2(x-1)^2+(y-1)^2=(\sqrt{2})^2

  1. Simplify the right-hand side. (2)2=2(\sqrt{2})^2 = 2, so:

(x−1)2+(y−1)2=2(x-1)^2+(y-1)^2=2

This is already a perfectly valid equation of the circle. Many exam questions accept this form.

  1. Expand if required (optional). Some problems ask for the general form x2+y2+Dx+Ey+F=0x^2+y^2+Dx+Ey+F=0. Let’s expand: …
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