Question

If the equation of two diameters of a circle are $2x+y=6$ and $3x+2y=4$ equation of circle.

Answer 1

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 equation of circle 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
=> $2\times 8+y=6$
=> $16+y=6$
=> $y=-10$
Here centre of circle $=(8,-10)$.

Equation of circle $=(x-h)²+(y-k)²=r²$.
where $(h,k)$ is the center of 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.