Question

If the points (3,0) (0,4) (0,0) and (K,4) are concyclic points then k=​

Answer 1

$\textbf{Given:}$

$\text{The points (3,0), (0,4), (0,0) and (k,4) are concyclic}$

$\textbf{To find:}$

$\text{The value of k}$

$\textbf{Solution:}$

$\text{Since the points are concyclic, they on a same circle}$

$\text{First we find out the equation of the circle which passes through}$

$(3,0),\;(0,4),\;\text{and}\;(0,0)$

$\text{Let the equation of the circle be}$

$x^2+y^2+2g\,x+2f\,y+c=0$

$\text{since the circle passes through (3,0), (0,4) and (0,0), we have}$

$\text{For (0,0)}$

$0^2+0^2+2g(0)+2f(0)+c=0$

$\implies\bf\,c=0$

$\text{For (3,0)}$

$3^2+0^2+2g(3)+2f(0)+0=0$

$9+6g=0$

$6g=-9$

$\implies\,g=\dfrac{-9}{6}$

$\implies\bf\,g=\dfrac{-3}{2}$

$\text{For (0,4)}$

$0^2+4^2+2g(0)+2f(4)+0=0$

$16+8f=0$

$8f=-16$

$\implies\bf\,f=-2$

$\text{The equation of the required circle is}$

$\bf\,x^2+y^2-3x-4y=0$

$\text{Since the circle also passes through (k,4), we have}$

$k^2+4^2-3k-4(4)=0$

$k^2+16-3k-16=0$

$k^2-3k=0$

$k(k-3)=0$

$\implies\,k=0,3$

$\text{For k=0, the corresponding point is (0,4)}$

$\therefore\textbf{The required value of k is 3}$