Question

Find the equation of a circle which has passing through (2,-3)(-4,5)which has the center on the 4x+3y+1=0​

Answer 1

Answer:

Let centre O(h,k),A(2,-3)and B(-4,5).

Let, OA=OB

So(h-2)^2+(k+3)^2=(h+4)^2+(k-5)^2

h^2+4-4h+k^2+9+6k = h^2+16+8h+k^2+25-10k

Further Simplification we get,

12h-16k+28=0-------------equation 1

It is given that the centre lies on the equation 4x+3y+1=0

So,(h,k) should be substituted in the equation 4x +3y+1=0

Then it would be 4h+3k+1=0

Here we got equation 1 is

12h-16k+28=0

So,To solve this equation we multiply 3 with the equation 4h+3k+1=0 then we get 12h+9k+3=0 which is equation 2

Now solve equation 1 and 2 we get

12h-16k+28=0

12h+9k+3=0

____________

-25k= -25

____________

So,k=1

Now substitute k value in equation 1 then we get

12h-16(1)+28=0

12h-16+28=0

12h+12=0

12h = -12

So , h= -1

We got (h,k)=(-1,1)

Now we should calculate OA= root over (x2-×1)^2+(y2-y1)^2

= root over 3^2+4^2

=root 25

=5

Now substitute (h,k) and r values in general equation of circle that is (x-h)^2+(y-k)^2=r^2

(x+1)^2+(y-1)^2=(5)^2

i.e, x^2+y^2+2x-2y-23=0 is the required equation of the circle.

Hope you like this answer if its lengthy also please read and understand the sum clearly and happily

Answer 2

Answer:

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