Question

if the equation of two diameters of a circle are 2x+y=6 and 3x+2y=4 and the radius is 10 , find the equation of the circle

Answer 1

hello friends....

given that
the equation of two diameters of a circle are 2x+y=6 and 3x+2y=4
and
the radius is 10 ,

we have to find
equation of circle

solution :-
solving 2x+y=6 ........(1)
and
3x+2y=4 .............(2)
we get to the center of circle

=> now multiple (1) by 2
we get
4x +2y = 12 ........(3)

now
subtract 2 from 3 we get
(4x +2y ) -(3x +2y) = 12-4
=> x = 8
putting the value of x in (1) we get
=] 2×8 +y = 6
=> 16+y = 6
=> y = -10

here
centre of circle = (8,-10)

now
we know that
equation of circle = (x-h)² + (y-k)² = r²
where (h,k) is the center of circle and r is the radius

=> (x-8)² + (y+10)² = 10²
=> x² +64 -16x +y² +100 +20y = 100
=> x² +y² -16x +20y +64=0

hence
x² +y² -16x +20y +64=0 is the required equation of circle

⭐⭐ hope it helps ⭐⭐

Answer 2

Answer:  Given the equation of two diameters of a circle are 2x+y=6 and 3x+2y=4 and  radius is 10 ,

we have to find an equation of circle solving 2x+y=6 .....(1) and 3x+2y=4 .......(2)

we get to the center of the circle

=> now multiple (1) by 2

we get 4x +2y = 12 ........(3)

Now subtract 2 from 3 we get

(4x+2y)−(3x+2y)=12−4

=> x=8

Putting the value of x in (1) we get

=> 2×8+y=6

=> 16+y=6

=> y=−10

Here centre of circle =(8,−10).  

Equation of circle =(x−h)²+(y−k)²=r².

where (h,k) is the center of the circle and r is the radius

=>(x−8)²+(y+10)²=10²

=> x²+64−16x+y²+100+20y=100

=> x²+y²−16x+20y+64=0

Hence x²+y²−16x+20y+64=0 is the required equation of the circle.

Hope this answer helps u...................