Question

Equation of circles passing through (3, -6) and touching both the axes is

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (D)

The correct options are
A

D

By looking at the question ,we can see that it is not easy to find the radius and centre of the circle.we

can't start with equation.

$(x-h{)}^2+(y-k{)}^2=r^2$

so we will start with the general equation of a circle and find g,f and c.

This makes sense because the conditions for touching the axes are in terms of g,f and c.

$x^2+y^2+2gx+2fy+c=0$ is the equation of circle.

we are given it touches both the axes .It means length of intercepts $2\sqrt{g^2-c}$ and $2\sqrt{f^2-c}$ is equal to

zero.

$\Rightarrow g^2=cand{f}^2=c$

$\Rightarrow g^2=f^2$

$f=\mp g$

so we can write the equation as $x^2+y^2+2gx\mp 2gy+g^2=0$

It passes through the point (3,-6)

$\Rightarrow 3^2+(-6{)}^2+2g\times 3\mp 2g\times (-6)+g^2=0$

$\Rightarrow 45+6g\mp 12g+g2=0$

$\Rightarrow g^2-12g+6g+45=0or{g}^2-12g+6g+45=0$

$\Rightarrow g^2-6g+45=0or{g}^2+18g+45=0$

$\Rightarrow (g-3{)}^2+36=0or(g+3\left)\right(g+15)=0$

$(g-3{)}^2+36$ can't be zero

$\Rightarrow$ no solution to $(g-3{)}^2+36=0$

$\Rightarrow g=-3,org=-5$

we will calculate f and c for these values.

$g=-3$ $g=-15$

$f=\mp 3$ $g=-15$

$c=(-3{)}^2=9$ $c=(-15{)}^2=225$

so the set of values of g,f and c are $(-3,\mp 3,9)$ and $(-15,\mp 15,225)$.

so the equations of circles are $x^2+y^2-6x\mp 6y+9=0$ and

$x^2+y^2-30x\mp 30y+225=0$

option A and D are correct.