Question

Choose the correct formula among the following:
lf $S=x^2+y^2+2gx+2fy+c=0$ represents the equation of a circle, then,
I. Length of the tangent drawn from $P(x_1,y_1)$ to the circle $S=0$ is $\sqrt{S_{11}}$
II. the equation of a circle with center $(r,y_1)$ and radius $r$ is $(x-r{)}^2+(y-y_1{)}^2=r^2$ then it touches $y$-axis at $(0,y_1)$
III. Length of the intercept made by the circle
on y-axis is $\sqrt{f^2-c}$

• A. Only I,II
• B. Only II,III
• C. Only I,III
• D. I,II,III

Answer 1

The correct option is A Only I,II

I.

$PA=$ length of tangent.
$=\sqrt{(Pc{)}^2-r^2}$
$=\sqrt{x_1^2+g^2+2g{x}_1+y_1^2+2f{y}_1+f^2-(g^2+f^2-c)}$
$=\sqrt{x_1^2+y_1^2+2g{x}_1+2f{y}_1+c}$
$\Rightarrow PA=\sqrt{S_{11}}$

Thus, it is true.

II.

By centre radius from of circle, having centre $(r,y_1)$ and radius $r$ is
$(x-r{)}^2+(y-y_1{)}^2=r^2$
$(0,y_1)$ lie on the circle.

We see that, circle touches Y-axis as

Distance between $(r,y_1)$ and $(0,y^1)$ is $r$ which is radius of circle.

III. For intercept on y-axis, put $x=0$
$y^2+2fy+c=0$
$|y_2-y_1|=\sqrt{(y_1+y_2{)}^2-4y_1{y}_2}$
$=\sqrt{(-2f{)}^2-4c}$
Intercept on Y-axis $=|y_2-y_1|=2\sqrt{f^2-c}$

Thus, it is false.