Question

If the centre $O$ of circle is the intersection of $x-$axis and line $y=\frac{4}{3}x+4$, and the point $(3,8)$ lies on circle, then the equation of circle will be

• A. $x^2+y^2=25$
• B. ${(x+3)}^2+y^2=25$
• C. ${(x+3)}^2+y^2=100$
• D. ${(x+3)}^2+{(y-8)}^2=100$

Answer 1

Answer: (C)

${(x+3)}^2+y^2=100$

The intersection of $x$-axis and the line $y=\frac{4x}{3}+4$ is $(-3,0)$, which is the centre of the circle.

Since the point $(3,8)$ lies on the circle, its radius becomes the distance between these points, which is $\sqrt{(3+3{)}^2+(8-0{)}^2}=10$

The equation thus becomes $(x+3{)}^2+(y-0{)}^2=(10{)}^2$

i.e. $(x+3{)}^2+y^2=100$