Question

Q4. A circle has radius 3 units and its centre lies on the line y = x - 1. Find the

equation of the circle ,if it passes through (7,3).​

Answer 1

Answer:

x ^2 +y ^2 −8x−6y+16=0

Step-by-step explanation:

(x−h) 2+(y−k) 2 =r 2

Here (h,k)→ centre

Equation of line through centre

k=h−1−−−−(1)

& (x,y) point on circle

So, given (x,y)=(7,3)

So, (7−h) 2+(3−h+1) 2 =3 2

Using eqn (1)

We get,

28+h 2 −11h=0

h=7 oR 4

k=6 oR 3

Equation of circle corresponding (7,6)

x 2 +y 2 −14x−12y+76=0

Equation of circle corresponding (4,3)

x 2 +y 2 −8x−6y+16=0

Answer 2

Step-by-step explanation:

${(x - h)}^{2} + {(y - k)}^{2} = {r}^{2}$

here (h,k) is centre

so , equation of line through centre

k = h - 1 --------(1)

so given (x,y) = (7,3)

so ,

${(7 - h)}^{2} + {(3 - h + 1)}^{2} = {3}^{2}$

using equation (1) we get

$28 + {h}^{2} - 11h = 0$

h=7 or 4

k= 6 or 3

equation of circle corresponding (7,6)

${x}^{2} + {y}^{2} - 14x - 12y + 76 = 0$

equation of circle corresponding (4,3)

${x}^{2} + {y}^{2} - 8x - 6x + 16 = 0$