Question

Find the equation of the circle , passing through the point (2 ,3) and ( -1,1) whose center on the line x-3y -11 =0​

Answer 1

Answer:

let the coordinates of centre of the circle be (h,k)since point(h,k) lie on the line

x-3y-11=0

therefore equation becomes

h-3y-11=0...............................eq1

the equation of the circle is (x-h)₂ +(y-k)₂=r₂..................eq₂

the circle passes through points(2,3) and (-1,1)

therefore substituting values of x and y in eq₂

(2-h)₂+(3-k)₂=r₂.................eq3

(-1-h)₂+(1-k)₂=r₂....................eq4

eq3=eq4

we get

6h+4k=11...................eq₅

solving equations 5 and 1 we get

h=7/2  , k= -5/2

put values of h and k in eq 3

we get r=√130/2

replace values of h, k and r in eq 2

the equation of circle comes out to be

(x-7/2)₂+(y+5/2)₂=130/4

 the student might have some problem in equation printing as i do not know how to use superscript and subscript....

Step-by-step explanation: