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Worked Examples · Example 10

Q.Find the equation of a circle with centre (3,−2)(3, -2) and radius 5.

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✓ Free question

Use the standard (centre-radius) form of a circle and expand it into general form.

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.

  1. Here h=3, k=−2, r=5h=3,\ k=-2,\ r=5. Write the standard form:

(x−3)2+(y+2)2=52=25(x-3)^2+(y+2)^2=5^2=25

  1. Expand each square: (x−3)2=x2−6x+9(x-3)^2=x^2-6x+9 and (y+2)2=y2+4y+4(y+2)^2=y^2+4y+4.
  2. Substitute: x2−6x+9+y2+4y+4=25x^2-6x+9+y^2+4y+4=25.
  3. Combine constants: x2+y2−6x+4y+13=25x^2+y^2-6x+4y+13=25.
  4. Move 25 to the left: x2+y2−6x+4y+13−25=0⇒x2+y2−6x+4y−12=0x^2+y^2-6x+4y+13-25=0\Rightarrow x^2+y^2-6x+4y-12=0.
  5. Self-check: centre =(−−62,−42)=(3,−2)=(-\tfrac{-6}{2},-\tfrac{4}{2})=(3,-2) ✓, radius =32+(−2)2−(−12)=9+4+12=25=5=\sqrt{3^2+(-2)^2-(-12)}=\sqrt{9+4+12}=\sqrt{25}=5 ✓.
✓Final answer

Equation of the circle: x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0.

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