Question

Let $\overrightarrow{r}$ be a position vector of a variable point $P$ in Cartesian plane. A tangent is drawn to the curve $\overrightarrow{r}\cdot (10\hat{j}-8\hat{i}-\overrightarrow{r})=-40$ from the point $(1,1).$ If $p_1=max\{|\overrightarrow{r}+2\hat{i}-3\hat{j}{|}^2\}$ and $p_2=min\{|\overrightarrow{r}+2\hat{i}-3\hat{j}{|}^2\},$ then which of the following is (are) CORRECT?

• A. $p_1=9+4\sqrt{2}$
• B. $p_1+p_2=18$
• C. Locus of $P$ represents a circle with centre $(4,-5)$
• D. Length of the tangent is $2\sqrt{10}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $p_1=9+4\sqrt{2}$
B $p_1+p_2=18$
D Length of the tangent is $2\sqrt{10}$

Let $\overrightarrow{r}=x\hat{i}+y\hat{j}$
$\overrightarrow{r}\cdot (10\hat{j}-8\hat{i}-\overrightarrow{r})=-40$
$\Rightarrow x^2+y^2+8x-10y+40=0$, which is a circle with centre $C(-4,5)$ and radius $r=1$


$p_1=max\left\{\right(x+2{)}^2+(y-3{)}^2\}$
$p_2=min\left\{\right(x+2{)}^2+(y-3{)}^2\}$
Let $P$ be $(-2,3)$.
Then $CP=2\sqrt{2},r=1$
The maximum and minimum value occur along the diameter.
$\therefore p_2=(CP-r{)}^2=(2\sqrt{2}-1{)}^2$
and $p_1=(CP+r{)}^2=(2\sqrt{2}+1{)}^2$
$\Rightarrow p_1+p_2=18$

We know that, length of tangent from an external point $(x_1,y_1)$ to the circle $x^2+y^2+2gx+2fy+c$ is $\sqrt{S_1}=\sqrt{x_1^2+y_1^2+2g{x}_1+2f{y}_1+c}$
Here, $(x_1,y_1)=(1,1)$
Length of tangent $=\sqrt{40}=2\sqrt{10}$