Question: Find the centre and radius of the following circles whose equations are:
[Remember these equations of circle:
x² + y² = r²
(x - h)² + (y - k)² = r²
(x - x1)(x - x2) + (y - y1)(y - y2) = 0]
Here,
h,g = x-coordinate of the centre
k,f = y-colrdinate of the centre
r = radius of the circle
a) x² + y² = 25
Solution:
Given equation of circle is x² + y² = 25
Above equation can also be written as:
(x - 0)² + (y -0)² = 5² - (i)
Comparing equation (i) with the (x - h)² + (y - k)² = r², we get,
h = 0, k = 0, r = 5
So, the centre of the circle is at (0,0) and the radius is 5 units.
c) (x +4)² + (y +5)² = 14
Solution:
Given equation of circle is (x +4)² + (y +5)² = 14
or, {x - (-4)}² + {y - (-5)}² = (√14)² - (i)
Comparing equation (i) with (x - h)² + (y - k)² = r², we get,
h = -4, k = -5, r = √14
So, the centre of the circle is at (-4,-5) and the radius is √14 units.
e) x² + y² - 18x + 71 = 0
Solution:
Given equation of circle is x² + y² - 18x + 71 = 0
or, x² + y² - 2×9x + 71 = 0 - (i)
Comparing equation (i) with x² + y² + 2gx + 2fy + c = 0, we get,
-g = 9, -f = 0, c = 71
Now,
r = $\sqrt{g² + f² - c }$
$or, r = \sqrt{(-9)² + 0 - 71}$
$or, r = \sqrt{81 - 71}$
$\therefore r = \sqrr{10} units$
Hence, the centre of the circle is at (9,0) and the radius is √10 units.
g) 2x - 6y - x² - y² = 1
Solution:
Given equation of circle is 2x -6y -x² -y² = 1
or, x² + y² -2x +6y +1 = 0
or, x² + y² - 2×1x + 2×3y +1 = 0 - (i)
Comparing equation (i) with x² + y² + 2gx + 2fy + c = 0, we get,
-g = 1, -f = -3, c = 1
Now,
r = $\sqrt{g² + f² - c}$
$or, r = \sqrt{1² + (-3)² - 1}$
$or, r = \sqrt{1 + 9 -1}$
$\therefore r = 3 units$
Hence, the centre of the circle is at (1,-3) and the radius is 3 units.
I) 5x² + 5y² - 50x + 20y - 65 = 0
Solution:
Given equation of circle is 5x² + 5y² - 50x + 20y - 65 = 0
or, 5(x² + y² - 10x + 4y -13) = 0
or, x² + y² - 10x +4y -13 = 0
or, x² + y² - 2×5x + 2×2y -13 = 0 - (i)
Comparing equation (i) with x² + y² + 2gx + 2fy + c = 0, we get,
-g = 5, -f = -2, c = -13
Now,
r = $\sqrt{g² + f² - c}$
$or, r = \sqrt{5² +(-2)² - (-13)}$
$or, r = \sqrt{25 + 4 +13}$
$\therefore r = \sqrt{42} units$
Hence, the centre of the circle is at (5,-2) and the radius is √42 units.
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