Question

The shortest distance from the point $P\left(-7,2\right)$ to the circle $x^2+y^2-10x-14y-151=0$ is in units

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Shortest distance from $P(-7,2)$ to $x^2+y^2-10x-14y-151=0$

$x^2+y^2-10x-14y-151=0$

$\Rightarrow (x-5{)}^2+(y-7{)}^2-151-25-49=0$

$\Rightarrow (x-5{)}^2+(y-7{)}^2=225$

$O=$centre$=(5,7)$

Radius$=\sqrt{225}$

$=15$

Shortest distance $=|OP-r|$

$=|\sqrt{(5+7{)}^2+(7-2{)}^2}-15|$

$=|\sqrt{{12}^2+5^2}-15|$

$=|\sqrt{169}-15|$

$=|13-15|$

$=2$

$\therefore$ Shortest distance from $P(-7,2)$ to circle $x^2+y^2-10x-14y-151=0$ is $2$ units.