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Exercise 12.2 · Q1

Q.Find the equation of the circle with:

(i) centre (0,2)(0, 2) and radius 2.
(ii) centre (0,0)(0, 0) and radius 3.
(iii) centre (−a,−b)(-a, -b) and radius a2−b2\sqrt{a^2 - b^2}.
Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
28% · 11/40 Questions
✓ Free question

Apply the standard circle equation (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2 to each given centre-radius pair and expand.

Circle with centre (h,k)(h,k) and radius rr: (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2.

(i) Centre (0,2)(0,2), radius 22:

  1. (x−0)2+(y−2)2=22⇒x2+y2−4y+4=4(x-0)^2+(y-2)^2=2^2\Rightarrow x^2+y^2-4y+4=4.
  2. Simplify: x2+y2−4y=0x^2+y^2-4y=0.

(ii) Centre (0,0)(0,0), radius 33:

3. (x−0)2+(y−0)2=32⇒x2+y2=9(x-0)^2+(y-0)^2=3^2\Rightarrow x^2+y^2=9.

(iii) Centre (−a,−b)(-a,-b), radius a2−b2\sqrt{a^2-b^2}:

4. (x−(−a))2+(y−(−b))2=(a2−b2)2⇒(x+a)2+(y+b)2=a2−b2(x-(-a))^2+(y-(-b))^2=\left(\sqrt{a^2-b^2}\right)^2\Rightarrow(x+a)^2+(y+b)^2=a^2-b^2.

5. Expand: x2+2ax+a2+y2+2by+b2=a2−b2x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2.

6. Simplify: x2+y2+2ax+2by+a2+b2−a2+b2=0⇒x2+y2+2ax+2by+2b2=0x^2+y^2+2ax+2by+a^2+b^2-a^2+b^2=0\Rightarrow x^2+y^2+2ax+2by+2b^2=0.

7. Self-check (i): centre from x2+y2−4y=0x^2+y^2-4y=0 is (0,2)(0,2), radius =0+4−0=2=\sqrt{0+4-0}=2 ✓.

✓Final answer

  1. x2+y2−4y=0x^2+y^2-4y=0
  2. x2+y2=9x^2+y^2=9
  3. x2+y2+2ax+2by+2b2=0x^2+y^2+2ax+2by+2b^2=0

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