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Worked Examples · Example 11

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

Ladakh CbseNCERTSubjective· 2mImportance★★★★★est
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✓ Free question

Find the radius as the distance from the given centre to the given point, then write the standard equation and expand.

Distance formula: d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. Circle: (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2.

  1. Centre (h,k)=(2,2)(h,k)=(2,2); the circle passes through (4,5)(4,5), so the radius is the distance between these points:

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}

  1. Standard form: (x−2)2+(y−2)2=13(x-2)^2+(y-2)^2=13.
  2. Expand: (x−2)2=x2−4x+4(x-2)^2=x^2-4x+4, (y−2)2=y2−4y+4(y-2)^2=y^2-4y+4.
  3. Substitute: x2−4x+4+y2−4y+4=13⇒x2+y2−4x−4y+8=13x^2-4x+4+y^2-4y+4=13\Rightarrow x^2+y^2-4x-4y+8=13.
  4. Rearrange: x2+y2−4x−4y+8−13=0⇒x2+y2−4x−4y−5=0x^2+y^2-4x-4y+8-13=0\Rightarrow x^2+y^2-4x-4y-5=0.
  5. Self-check with (4,5)(4,5): 16+25−16−20−5=0⇒41−41=016+25-16-20-5=0\Rightarrow41-41=0 ✓.
✓Final answer

Equation of the circle: x2+y2−4x−4y−5=0x^2+y^2-4x-4y-5=0 (radius 13\sqrt{13} units).

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