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The endpoints of a diameter of a circle are (−5, −2) and (−2, −6). Find the co-ordinates of the centre and radius of the circle. - Mathematics

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Question

The endpoints of a diameter of a circle are (−5, −2) and (−2, −6). Find the co-ordinates of the centre and radius of the circle.

Sum
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Solution

Formulas:

I. Midpoint (centre of a diameter):

Midpoint of (x1, y1) and (x2, y2) is

m = `((x_1 + x_2) / 2, (y_1 + y_2) / 2)`

II. Distance between two points (length of diameter):

d = `sqrt[(x_2 − x_1)^2 + (y_2 − y_1)^2]`

III. Radius: r = `("diameter")/2`

IV. Equation of a circle with centre (h, k) and radius r:

`(x − h)^2 + (y − k)^2 = r^2`

Given endpoints of the diameter A(−5, −2) and B(−2, −6):

⇒ Centre (midpoint of AB) h;

m = `(−5 + (−2)) / 2, (−2 + (−6)) / 2`

= `(−7) / 2, (−8) / 2`

∴ m = `(−7/2, −4)`

Centre =`(−7/2, −4)`

⇒ Diameter length (distance AB) AB;

d = `sqrt[(−2 − (−5))^2 + (−6 − (−2))^2]`

= `sqrt[(3)^2 + (−4)^2]`

= `sqrt[9 + 16]`

= `sqrt(25)`

∴ d = 5

⇒ Radius r = `AB / 2`

= `5/2`

∴ r = 2.5

Hence, circle equation Centre (h, k) = `(−7/2, −4),

`r^2 = (5/2)^2`

= `25/4`

So,

`(x + 7/2)^2 + (y + 4)^2 = 25/4`

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Chapter 11: Section formula - Exercise 11A [Page 229]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 11 Section formula
Exercise 11A | Q 21. | Page 229
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