Question

I. If the points $(a,0)$ , $(b,0)$ , $(0,c)$ , $(0,d)$ are concyclic, then $ab=cd$
II. If the points $(1,-6),(5,2),(7,0),(-1,k)$ are concyclic then $k=-3$.

• A. Only I is true
• B. Only II is true
• C. I & II are true
• D. Neither I & II are true

Answer 1

The correct option is A Only I is true

I) Let equation of circle passing through $(a,0),(b,0),(0,c),(0,d)$ is

$x^2+y^2+2gx+2fy+K=0$

Put (a,0) $a^2+2ga+K=0----\left(1\right)$

Put (b,0) $b^2+2gb+K=0----\left(2\right)$

From (1) & (2) $a^2-b^2+2g(a-b)=0$

K=-ab and $g=\frac{-(a+b)}{2a}$

and put $(0,c)$

$c^2+2fc+K=0----\left(4\right)$

and put $(0,d)$

$d^2+2fd+K=0----\left(5\right)$

$f=-\left(\right(k+d)$

From (4)& (5),

So, $K=-cd$ ---- (6)

From (3) & (6),

$K=-ab=-cd$

$ab=cd.$ True

ii) let $x^2+y^2+2gx+2fy+c=0$

put $(1,-6),(5,2),(7,0)$ and $(-2,k)$ in $e{q}^n$ of circle and solve then get value of $K$