Question

$x^2+y^2-14x-10y+24=0,$ makes an

Answer 1

A)Substituting $y=0$ in equation of circle we get $x^2-14x+24\Rightarrow (x-12)(x-2)=0$
Hence intercept on circle is $=12-2=10$
B) Substituting $x=0$ in equation of circle we get $y^2-10y+24=0\Rightarrow (y-6)(y-4)=0$
Hence intercept on y-axis is $=6-4=2$
C) Substituting $y=x$ in equation of circle we get $2x^2-24x+24=0\Rightarrow x=y=6\pm 2\sqrt{6}$
Hence intercept on $y=x$ line is $=\sqrt{(2\times 2\sqrt{6}{)}^2+(2\times 2\sqrt{6}{)}^2}=8\sqrt{3}$
D) Substituting y from $7x+y-4=0$ in equation of circle we get $x^2+(4-7x{)}^2-14x-10(4-7x)+24=0\Rightarrow 50x^2=0\Rightarrow x=0,y=4$
Hence intercept of $7x+y-4=0$ is 0