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Exercise 10.1 · Q5

Q.Find the equation of the circle with centre (−a,−b)(-a, -b) and radius a2−b2\sqrt{a^2 - b^2}.

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A circle with centre (h,k)(h, k) and radius rr has equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. Substituting centre (−a,−b)(-a, -b) and radius a2−b2\sqrt{a^2 - b^2} gives (x+a)2+(y+b)2=a2−b2(x + a)^2 + (y + b)^2 = a^2 - b^2.

The standard form of a circle's equation captures a beautiful geometric idea: every point on the circle is exactly the same distance from the centre. If the centre is at (h,k)(h, k) and the radius is rr, then any point (x,y)(x, y) on the circle satisfies the distance formula

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

Squaring both sides removes the square root and gives us the cleaner algebraic form:

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

This is the standard equation of a circle. Now we simply substitute the given values.

Step-by-step substitution:

  1. Identify the centre coordinates. We're told the centre is at (−a,−b)(-a, -b), so h=−ah = -a and k=−bk = -b.

  2. Identify the radius. The radius is r=a2−b2r = \sqrt{a^2 - b^2}.

  3. Square the radius. Since the standard form requires r2r^2, we compute:

    r2=(a2−b2)2=a2−b2r^2 = \left(\sqrt{a^2 - b^2}\right)^2 = a^2 - b^2 …

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