Question

Find the equation of the circle which passes through the points (5, 0) and (1, 4) and whose centre lies on the line x + y – 3 = 0.

Answer 1

long solution but hope you'll understand.... :)

Answer 2

Answer:

$(x-2)^2+(y-1)^2=10$

Step-by-step explanation:

Equation of circle : $(x-h)^2+(y-k)^2=r^2$  --A

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

We are given that center lies on the line x+y-3=0

So, Substitute (h,k)

h+k-3=0

h+k=3  --1

We are given that Circle passes through the points (5,0) and (1,4)

Substitute (5,0) in A

$(5-h)^2+(0-k)^2=r^2$

$25+h^2-10h+k^2=r^2$

$h^2-10h+k^2+25=r^2$  --2

Substitute (1,4) in A

$(1-h)^2+(4-k)^2=r^2$  

$1+h^2-2h+16+k^2-8k=r^2$  

$h^2-2h+k^2-8k+17=r^2$  --3

Subtract 3 from 2

$h^2-10h+k^2+25-h^2+2h-k^2+8k-17=r^2-r^2$

$-10h+25+2h+8k-17=0$

$-8h+8k+8=0$

$-h+k+1=0$

$h-k=1$   ---4

Add 1 and 4

$h-k+h+k=1+3$

$2h=4$

$h=2$

Substitute value of h in 4

$2-k=1$

$k=1$

Substitute value of h and k in 2

$(2)^2-10(2)+1^2+25=r^2$  

$4-20+1+25=r^2$  

$10=r^2$  

Substitute the value of h , k and $r^2$ in A

$(x-2)^2+(y-1)^2=10$

So, the equation of the circle is $(x-2)^2+(y-1)^2=10$