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Exercise 10.1 · Q7

Q.Find the centre and radius of the circle x2+y2−4x−8y−45=0x^2 + y^2 - 4x - 8y - 45 = 0.

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The equation is rewritten by completing the square to get (x−2)2+(y−4)2=65(x-2)^2 + (y-4)^2 = 65, so the centre is (2,4)(2,4) and the radius is 65\sqrt{65}.

The standard form of a circle’s equation is (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, where (h,k)(h,k) is the centre and rr is the radius. Our given equation is in expanded form — it has x2x^2, y2y^2, linear terms in xx and yy, and a constant. To extract the centre and radius, we need to reverse the expansion: that is, complete the square for the xx terms and the yy terms separately.

Why does completing the square work? Because a perfect square like (x−h)2(x - h)^2 expands to x2−2hx+h2x^2 - 2hx + h^2. If we have x2−4xx^2 - 4x, we can ask: what constant h2h^2 would make this a perfect square? Here −2h=−4-2h = -4, so h=2h = 2 and h2=4h^2 = 4. So x2−4x=(x−2)2−4x^2 - 4x = (x-2)^2 - 4. We do the same for yy.

Let’s go step by step.

  1. Group the xx and yy terms Write the equation as:

(x2−4x)+(y2−8y)=45(x^2 - 4x) + (y^2 - 8y) = 45

We moved the constant −45-45 to the right side as +45+45.

  1. Complete the square for xx For x2−4xx^2 - 4x, take half of −4-4, which is −2-2, square it to get 44. Then:

x2−4x=(x−2)2−4x^2 - 4x = (x-2)^2 - 4

  1. Complete the square for yy For y2−8yy^2 - 8y, half of −8-8 is −4-4, square it to get 1616. Then:

y2−8y=(y−4)2−16y^2 - 8y = (y-4)^2 - 16

  1. Substitute back into the equation

(x−2)2−4+(y−4)2−16=45(x-2)^2 - 4 + (y-4)^2 - 16 = 45

Combine the constants: −4−16=−20-4 - 16 = -20, so:

(x−2)2+(y−4)2−20=45(x-2)^2 + (y-4)^2 - 20 = 45

  1. Isolate the squared terms Add 2020 to both sides: (x−2)2+(y−4)2=65(x-2)^2 + (y-4)^2 = 65 …
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←Previous question · 10.3Q6 — Find the centre and radius of the circle (x + 5)² + (y - 3)² = 36.→Next question · 10.3Q8 — Find the centre and radius of the circle x² + y² - 8x + 10y - 12 = 0.