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# Find the coordinates of the point on the circle x²+y²-12x-4y+30=0 which is nearest the origin.?

Ans: (3,1)

### 2 Answers

- Φ² = Φ+1Lv 72 years agoFavorite Answer
x²+y²-12x-4y+30=0 is a circle centred at (6,2) with a radius of √(6²+2²-30) = √10

and (6,2) is √(6²+2²) = 2√10 from the origin {twice the length of the radius!}

So the point on the circle x²+y²-12x-4y+30=0 which is nearest the origin is the midpoint of (0,0) and (6,2) which is (3,1).

- 2 years ago
Find the center of the circle

x^2 - 12x + y^2 - 4y = -30

x^2 - 12x + 36 + y^2 - 4y + 4 = -30 + 36 + 4

(x - 6)^2 + (y - 2)^2 = 10

The center is at (6 , 2)

Describe a line that passes through (0 , 0) and (6 , 2)

m = (2 - 0) / (6 - 0) = 2/6 = 1/3

y = (1/3) * x

Find when this intersects the circle

x^2 + (x/3)^2 - 12x - 4 * (x/3) + 30 = 0

x^2 + (1/9) * x^2 - 12x - 4x/3 + 30 = 0

(10/9) * x^2 - (12 + 4/3) * x + 30 = 0

10 * x^2 - 9 * (12 + 4/3) * x + 9 * 30 = 0

10x^2 - (108 + 12) * x + 270 = 0

10x^2 - 120x + 270 = 0

x^2 - 12x + 27 = 0

x^2 - 12x = -27

x^2 - 12x + 36 = -27 + 36

(x - 6)^2 = 9

x - 6 = -3 , 3

x = 3 , 9

x = 3

y = x/3

y = 3/3

y = 1

(3 , 1)