Question

find the equation of the circle with diameter ab of(-3,2) and b(1,4)

Answer 1

Given

coordinate of Diameter of circle

To Find

We have to find out the equation of circle

Solution :-

Formula :-

$\boxed{\sf\ (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0}$

We have ,

$\sf\ \ x_1= (-3) \ \ ;\ \ y_1=2\\ \\ \sf\ \ x_2=1\ \ ;\ \ y_2=4$

Put these values in the formula -

$:\implies\sf\ [x-(-3)](x-1)+(y-2)(y-4)=0\\ \\ \\ :\implies\sf\ (x^2-x+3x-3)+(y^2-4y-2y+8)=0\\ \\ \\ :\implies\sf\ x^2+2x-3+y^2-6y+8=0\\ \\ \\ :\implies\sf\ x^2+y^2+2x-6y+5=0$

$\underline{\boxed{\sf\ \ x^2+y^2+2x-6y+5=0}}$

Answer 2

AB is diameter ⟹ centre is midpoint of AB which is (

2

−1+3

,

2

2+3

)=(1,2.5)

AB=

(−1−3)

2

+(2−3)

2

=

17

⟹ radius =

2

AB

=

2

17

Equation of circle is (x−1)

2

+(y−2.5)

2

=

4

17

=4.25

# Answer is correct