Question

The nearest point of the circle $x^2+y^2-6x+4y-12=0$ from $\left(-5,4\right)$ is

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

Answer 1

Answer: (B)

(-1,1)

The equation of circle is,

$x^2+y^2-6x+4y-12=0$

$\therefore (x^2-6x)+(y^2+4y)-12=0$

$\therefore (x^2-6x+9)-9+(y^2+4y+4)-4-12=0$

$\therefore {(x-3)}^2+{(y+2)}^2-25=0$

$\therefore {(x-3)}^2+{[y-(-2)]}^2={\left(5\right)}^2$

Compare with standard form of equation of circle i.e. ${(x-h)}^2+{[y-k]}^2=r^2$, we get,

$h=3$, $k=-2$ and $r=5$

As shown in figure, coordinates of C are $C(3,-2)$, and $AC=5$

Let coordinates of point D are $(-5,4)$

As shown in figure, shortest distance between point D and circle is AD.

Now, $AD=CD-AC$ Equation (1)

By distance formula, $CD=\sqrt{{(3-(-5))}^2+{(-2-4)}^2}$

$\therefore CD=\sqrt{{\left(8\right)}^2+{(-6)}^2}$

$\therefore CD=\sqrt{64+36}=\sqrt{100}$

$\therefore CD=10$

From equation (1),

$AD=10-5$

$\therefore AD=5$

Now, Let coordinates of point A are $A(h,k)$

$\frac{CA}{AD}=\frac{5}{5}=1$

$\therefore \frac{m}{n}=\frac{1}{1}$

$\therefore m=1$ and $n=1$

Point A is dividing segment CD internally. Thus, by section formula,

$h=\frac{m{x}_2+n{x}_1}{m+n}$

$\therefore h=\frac{(1\times -5)+(1\times 3)}{1+1}$

$\therefore h=\frac{-5+3}{2}$

$\therefore h=-1$

Similarly, $k=\frac{m{y}_2+n{y}_1}{m+n}$

$\therefore k=\frac{(1\times 4)+(1\times -2)}{1+1}$

$\therefore k=\frac{4-2}{2}$

$\therefore k=1$

Thus, coordinates of point A are $A(-1,1)$

Thus, answer is option (B)