A circle has a radius of 3 units and its centre lies on the line y = x - 1. Find the equation of the circle if it passes through point (7, 3).

Correct option is B

Given:

A circle has a radius of 3 units 

circle centre lies on the line y = x - 1

circle passes through = 7,3

Formula Used:

(x−h)$$^2$$​+(y−k)$$^2$$​=r$$^2$$ 

Solution:

general equation of circle when h and k are co ordinate of circle 

(x−h)$$^2$$​​+(y−k)$$^2$$​​=r$$^2$$

put x= 7 and y =3 and r = 3 because radius =3

k=h-1 value putting of x and y in the line equation.

​$$(x - h)^2 + (y - k)^2 = r^2 \\r = 3 \\k = h - 1 \\(7 - h)^2 + (3 - k)^2 = 9 \\(7 - h)^2 + (4 - h)^2 = 9 \\(7 - h)^2 = 49 - 14h + h^2 \\(4 - h)^2 = 16 - 8h + h^2 \\49 + 16 - 14h - 8h + 2h^2 = 9 \\56 - 22h + 2h^2 = 0 \\h^2 - 11h + 28 = 0 \\h = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(28)}}{2(1)} \\h = \frac{11 \pm \sqrt{121 - 112}}{2} \\h = \frac{11 \pm \sqrt{9}}{2} \\h = \frac{11 + 3}{2} = 7 \quad \text{or} \quad h = \frac{11 - 3}{2} = 4 \\k = h - 1 \\(x - 7)^2 + (y - 6)^2 = 9 \\(x - 4)^2 + (y - 3)^2 = 9$$ 

we choose h=4 according to given option

​$$(x - 4)^2 + (y - 3)^2 = 9 \\(x - 4)(x - 4) + (y - 3)(y - 3) = 9 \\x^2 - 8x + 16 + y^2 - 6y + 9 = 9 \\x^2 + y^2 - 8x - 6y + 25 = 9 \\x^2 + y^2 - 8x - 6y + 25 - 9 = 0 \\x^2 + y^2 - 8x - 6y + 16 = 0$$​​

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