Question

If end points of diameter of a circle are $(12,-1)$ and $(-4,3)$ then the intercept made by it on $y$ -axis is:

• A. 2$\sqrt{13}$
• B. $\sqrt{13}$
• C. 4$\sqrt{13}$
• D. None

Answer 1

Answer: (C)

4$\sqrt{13}$

End points are $(12,-1)$ & $(-4,3)$

so centre is $(4,1)$

radius is $\sqrt{(12-4{)}^2+(-1-1{)}^2}$

$=\sqrt{64+4}=\sqrt{68}$

equation of circle:

$(x-4{)}^2+(y-1{)}^2=68$

y-axis means $x=0$

so, $16+(y-1{)}^2=68$

$(y-1{)}^2=52$

$y-1=\pm 2\sqrt{13}$

$y=2\sqrt{13}+1,-2\sqrt{13}+1$

Intercept$=2\sqrt{13}+1-(-2\sqrt{13}+1)$$=4\sqrt{13}$.