Question

find the equation of a circle passing through (4, 1) and (6, 5 )and having centre on the line 4 x + y - 16 =0

Answer 1

So, assuming the equation of the REQUIRED CIRCLE to be (x-h)²+(y-k)²=r²
We know that circle passes through the points (4,1) and (6,5)
(4-h)²+(1-k)²=r².......(1)
(6-h)²+(5-k)²=r²........(2)

The centre (h,k) of the circle lies on the line 4x+y-16=0
so, 4x+y=16.......(3)
From eq. (1) and (2), we get
(4-h)²+(1-k)²=(6-h)²+(5-k)²
⇒16-8h+h²+1-2k+k²=36-12h+h²+25-10k+k²
⇒16-8h+1-2k=36-12h+25-10k
⇒4h+8k=44
⇒h+2k=11....(4)

so, from eq. (3) and (4), we get h=3 and k=4

Now, substitute the values of h and k in eq. (1)
(4-3)²+(1-4)²=r²
⇒(1)²+(-3)²=r²
⇒1+9=r²
r²=10
r=√10

hence the equation of the required circle is
(x-3)²+(y-4)²=(√10)²
x²-6x+9+y²-8y+16=10
x²+y²-6x-8y+15=0

Hope this helps..!!