Question

$(x-1)(y-2)=5$ and $(x-1{)}^2+(y+2{)}^2=r^2$ intersect at four points $A,B,C,D$ and if centroid of $\Delta ABC$ lies on line $y=3x-4$, then locus of $D$ is

• A. $y=3x$
• B. $x^2+y^2+3x+1=0$
• C. $3y=x+1$
• D. $y=3x+1$

Answer 1

The correct option is A $y=3x$

Given, $(x-1)(y-2)=5$
Let $x-1=X,y-2=Y$
$X^2+{(Y+4)}^2=r^2$ .......(1)
$XY=5$ .......(2)
Substitute $X$ from (2) to (1), we get
$\frac{25}{Y^2}+Y^2+8Y+16=r^2$
$Y^4+8Y^3+{16Y}^2-r^2{Y}^2+25=0$
Sum of the roots $Y_1+Y_2+Y_3+Y_3=-8$
$y_1+y_2+y_3+y_3=0$
If $(x_i,y_i)$ is the point of intersection of given curves, then
$\frac{\underset{j=1}{\sum ^4{x}_i}}{4}=\frac{1+1}{2}$ and $\frac{\underset{j=1}{\sum ^4{y}_i}}{4}=0$
Now, $\frac{\underset{i=1}{\sum ^3{x}_i}}{3}=\frac{4-x_4}{3}$ and $\frac{\underset{i=1}{\sum ^3{y}_i}}{3}=-\frac{y_4}{3}$
Centroid $(\frac{{\sum }_{j=1}^3{x}_1}{3},\frac{{\sum }_{i=1}^3{y}_1}{3})$ lies on the line $y=3x-4$
Hence, $\frac{-y_4}{3}=\frac{3(4-x_4)}{3}-4$
$\Rightarrow y_4=3x_4$
Therefore, the locus of $D$ is $y=3x$.