Question

If $A(-2,3)B(2,-1)C(3,1)$ then the locus of $P$ such that $P{A}^2+P{B}^2+P{C}^2=7$ is a circle with centre

• A. $(2,1)$
• B. $(1,2)$
• C. $(1,1)$
• D. $(2,2)$

Answer 1

The correct option is C $(1,1)$

A$(-2,3)$,B$(2,-1)$,C$(3,1)$ then p$(x,y)$ so,

$P{A}^2+P{B}^2+P{C}^2=7$

$(x+2{)}^2+(y-3{)}^2+(x-2{)}^2+(y+1{)}^2+(x-3{)}^2+(y-1{)}^2=7$

$3x^2+3y^2-6x-6y+36=7$

$3x^2+3y^2-6x-6y+29=0$

Equation of circle in General form:

$x^2+y^2+2gx+2fy+c=0$, Where $(–g,–f)$ is a centre.

so, locus of p$(x,y)$ is circle with centre $(1,1)$