Question

The length of intercepts made by the circle $x^2+y^2-4x+6y+4=0$ on $X$ and $Y$ axis respectively, are

• A. $0,2\sqrt{5}$
• B. $2\sqrt{5},0$
• C. $0,0$
• D. $0,2$

Answer 1

The correct option is A $0,2\sqrt{5}$

Given equation: $x^2+y^2-4x+6y+4=0$
Comparing with general form of circle, so we get
$g=-2,f=3,c=4$

Now, the intercept made on $X-\text{axis}$
$=2\sqrt{g^2-c}=2\sqrt{(-2{)}^2-4}=0$
The intercept made on $Y-\text{axis}$
$=2\sqrt{f^2-c}=2\sqrt{9-4}=2\sqrt{5}$

Alternate method:
Given equation: $x^2+y^2-4x+6y+4=0$
Putting $x=0$
$y^2+6y+4=0$
It intersect $y-$ axis at $y_1,y_2$, so the $y-$ intercept
$=|y_1-y_2|=\sqrt{(y_1+y_2{)}^2-4y_1{y}_2}=\sqrt{36-16}=2\sqrt{5}$

Putting $y=0$
$x^2-4x+4=0\Rightarrow (x-2{)}^2=0$
So, $x-$ intercept is $0.$