Question

The centre of a circle is $(2a+3,2a-1)$. Find the value of $a$ if the circle passes through the point $(11,9)$ and has diameter of length $20$ units.

Answer 1

Center of circle $=(2a+3,2a-1)$

Point of which circle Pass $=(11,9)$

Radius of circle $=10$ units

Now $OA=\sqrt{(11-20-3{)}^2+(9-{20}^o+1{)}^2}$

Squaring both sides we get

$(10{)}^2=(8-2a{)}^2+(10-2a{)}^2$

$100=64+{40}^{2}032a+100+{40}^2-40a$

${80}^2-72a+64=0$

$a^2-9a+8=0$

$(a-8)(a-1)=0$

$[a=8$ or $a=1]$.