Question

In the $x-y$,the segment with end points $(3,8)$ and $(-5,2)$ is the diameter of the circle. The point $(k,10)$ lies on the circle for ?

• A. no value of k
• B. exactly one integral k
• C. exactly one non integral k
• D. two real value of k

Answer 1

The correct option is A no value of k

If $(x_1,y_1)$ and $(x_2,y_2)$ are the endpoints of the diameter of a circle, then the equation of the circle is given by:
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
Substituting the values given in the question, the circle is given by:
$(x-3)(x+5)+(y-8)(y-2)=0$
Substituting the point $(k,10)$ in the equation:
$(k-3)(k-5)+\left(2\right)\left(8\right)=0$
$⟹k^2-8k+15+16=0$
$⟹k^2-8k+31=0$
Discriminant of the above equation: $8^2-4\cdot 31<0$
Therefore, no value of k.