Question

$A=(2,1)$ Locus of B is $x^2+y^2=25.$ If P divides AB in the ratio $1:2$ then locus of p is

• A. $(3x-4{)}^2+(3y-2{)}^2=25$
• B. $(3x+1{)}^2+(3y-2{)}^2=25$
• C. $(x-3{)}^2+(y-2{)}^2=4$
• D. $(5x-2{)}^2+(5y+2{)}^2=25$

Answer 1

The correct option is A $(3x-4{)}^2+(3y-2{)}^2=25$

Let P be $(h,k)$ and $B(x,y)$ and $A(2,1)$

Ratio is $1:2$

$(h,k)=(\frac{2\left(2\right)+x}{2+1},\frac{2\left(1\right)+y}{2+1})=>(x,y)=(3h-4,k(3)-2)$

$(x,y)$ lies on its locus =>${(3h-4)}^2+{(3k-2)}^2=25$

For locus=>${(3x-4)}^2+{(3y-2)}^2=25$