Question

The number of circles passes through (3,-6) and touches both the axes are

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

0

Equation of circle touches both the axes is given by,
$(x-a{)}^2+(y-a{)}^2=a^2$
Given it also passes through $(3,-6)$
$\Rightarrow (3-a{)}^2+(a+6{)}^2=a^2\Rightarrow a^2+6a+45=0$
Clearly this quadratic does not posses real roots.Hence no circle is possible.