On a wheel there are three points (5, 7), (-1, 7) and (5, -1) located such that the distance from a fixed point to these points is always equal. Find Coordinate of the fixed point and then derive equation representing the locus that contains all three points. please solve step by step . Created by sakshiroyamkh.

On a wheel there are three points (5, 7), (-1, 7) and (5, -1) located such that the distance from

Answer:

Given figure is a circle.

Let the coordinate of the fixed point (centre) be (h,k) and the distance between this fixed point and the other points on the locus (radius) be represented by r.

Equation of circle is (x - h)² + (y - k)² = r²

Given, points on the locus (x,y) are (5,7), (-1,7) and (5,-1).

Since, the circle passes through (5,7), (-1,7) and (5,-1), we get,

i) (5 - h)² + (7 - k)² = r²

ii) (-1 - h)² + (7 - k)² = r²

iii) (5 - h)² + (-1 - k)² = r²

Put value of r² from equation (ii) in equation (i), we get,

(5 - h)² + (7 - k)² = (-1 - h)² + (7 - k)²

or, (5 - h)² - (-1 - h)² = (7 - k)² - (7 - k)²

or, (25 - 10h + h²) - (1 + 2h + h²) = 0

or, 12h - 24 = 0

or, 12h = 24

So, h = 2

Put value of r² from equation (iii) in equation (i), we get,

(5 - h)² + (7 - k)² = (5 - h)² + (-1 - k)²

or, (5 - h)² - (5 - h)² = (-1 - k)² - (7 - k)²

or, 0 = (1 + 2k + k²) - (49 - 14k + k²)

or, 0 = -48 + 16k

or, 16k = 48

So, k = 3

So, centre (h,k) = (2,3)

Take (x,y) = (5,7) and (h,k) = (2,3), we have,

(5 - 2)² + (3 - 7)² = r²

or, 9 + 16 = r²

So, r = 5

And,

Equation of circle when centre (h,k) = (1,2) and radius (r) = 5 is

(x - h)² + (y - k)² = r²

or, (x - 2)² + (y -3)² = (5)²

or, (x² - 4x + 4) + (y² - 6y + 9) = 25

or, x² + y² - 4x - 6y + 13 - 25 = 0

or, x² + y² - 4x - 6y -12 = 0 is the required equation.

Hence, the required equation of the given circle is x² + y² - 4x - 6y - 12 = 0.

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