Question

A circle touches x-axis at +3+3 in +ve direction of y-axis. Its equation is -

• A. $x^2+y^2+6x+10y-9=0$
• B. $x^2+y^2-6x-10y-9=0$
• C. $x^2+y^2-6x-10y+9=0$
• D. $x^2+y^2+6x+10y+9=0$

Answer 1

The correct option is C $x^2+y^2-6x-10y+9=0$

Given that a circle touches$+3$ distance at x-axis and $+8$ unit distance at y-axis. $R$ is the radius.

Consider the diagram shown below.

Since, the circle touches the x-axis at $(3,0)$, the x-coordinate of the centre will be $3$.

Also, intercept at y-axis $=8$

Therefore,

$OS=8$

$OH=\frac{8}{2}=4$

$\Rightarrow CH=3$ and $\Rightarrow OH=4$

Using Pythagoras theorem, we have

$OC=\sqrt{3^2+4^2}=5$

So,

$CQ=5$

Therefore,

Centre of the circle will be at $(3,5)$

Now,

${(x-3)}^2+{(y-5)}^2=5^2$

$x^2+9-6x+y^2+25-10y=25$

$x^2+y^2-6y-10y+9=0$

Hence, this is the required equation of the circle.