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NCERT Exemplar · Q8

Q.Find the equation of the circle having (1,−2)(1, -2) as its centre and passing through 3x+y=143x + y = 14, 2x+5y=182x + 5y = 18.

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The circle's centre is given; we find the radius by computing the perpendicular distance from the centre to the point of intersection of the two lines, then write the standard equation. The equation is (x−1)2+(y+2)2=25(x - 1)^2 + (y + 2)^2 = 25.

The standard form of a circle with centre (h,k)(h, k) and radius rr is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. We already know the centre (1,−2)(1, -2), so the equation will be (x−1)2+(y+2)2=r2(x - 1)^2 + (y + 2)^2 = r^2. The phrase "passing through 3x+y=143x + y = 14, 2x+5y=182x + 5y = 18" means the circle passes through the point where these two lines intersect. Once we find that point, the radius is simply the distance from the centre to it.

Finding the point of intersection

The two lines are:

3x+y=14…(i)3x + y = 14 \quad \text{…(i)}

2x+5y=18…(ii)2x + 5y = 18 \quad \text{…(ii)}

  1. Eliminate one variable. Multiply equation (i) by 5:

15x+5y=70…(iii)15x + 5y = 70 \quad \text{…(iii)}

  1. Subtract equation (ii) from (iii):

(15x+5y)−(2x+5y)=70−18(15x + 5y) - (2x + 5y) = 70 - 18

13x=5213x = 52

x=4x = 4

  1. Substitute x=4x = 4 into equation (i):

3(4)+y=143(4) + y = 14

12+y=1412 + y = 14

y=2y = 2

The two lines intersect at (4,2)(4, 2).

Computing the radius

The radius is the distance from the centre (1,−2)(1, -2) to the point (4,2)(4, 2):

r=(4−1)2+(2−(−2))2=32+42=9+16=25=5r = \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 …

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