Question

If (3,-2) is the centre of the circle $S= x^2+y^2+2gx+2fy-23=0$ and A is a point on the circle S = 0 such that its distance from a point P(-1,-5) is least, then A =

• A. $(3,-2)$
• B. $(9/5, 28/5)$
• C. $(3/5, 2/5)$
• D. $(-9/5, -28/5)$

Hint:

To find the point on a circle closest to or farthest from a given point P, draw a line through the center C and P. The intersections of this line with the circle are the required points. If P is inside the circle, the closest point is on the line segment CP extended away from C, and the farthest point is on the line segment PC extended away from P.

Answer 1

The Correct Option is D

The centre of the circle $x^2+y^2+2gx+2fy+c=0$ is $(-g, -f)$.
We are given that the centre is $C(3,-2)$. So, $-g=3 \implies g=-3$ and $-f=-2 \implies f=2$.
The equation of the circle is $x^2+y^2-6x+4y-23=0$.
The radius of the circle is $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+2^2-(-23)} = \sqrt{9+4+23} = \sqrt{36}=6$.
We need to find a point A on the circle that has the least distance from the point P(-1,-5).
The point A lies on the line segment connecting the centre C and the point P. It is the intersection of the line segment CP and the circle, closer to P.
The coordinates of the centre are $C(3, -2)$ and the external point is $P(-1, -5)$.
Point A divides the line segment CP internally in the ratio $r:d-r$, where $d$ is the distance CP. A simpler way is to find the point that divides CP externally in the ratio $d:r$ from P's perspective or internally in the ratio $(d-r):r$ from C's perspective. The point of least distance A divides CP internally in the ratio $r: (d-r)$, but it is easier to think of it as A divides the segment PC in the ratio $r: (d-r)$. This is complicated.
A simpler approach: A divides the line segment PC in the ratio $(d-r):r$. No, that's not right. A divides CP in some ratio. Let's find the ratio. A is on the circle. C is the center. P is outside. The closest point A is on the line segment CP. The ratio CA:AP is required. We know CA = radius = 6. Let's find the distance CP.
$d = CP = \sqrt{(3 - (-1))^2 + (-2 - (-5))^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25}=5$.
Wait, the distance from the centre to P is 5, which is less than the radius 6. This means the point P(-1,-5) is inside the circle.
If P is inside the circle, the point A on the circle with the least distance to P is on the line extending from C through P to the circle.
Point A divides the segment PC externally. The ratio is $CA:AP = 6:1$. So A divides the segment CP in the ratio 6:-1 is not right.
Let's use the section formula. Let A be $(x,y)$. A divides the segment CP in the ratio $C-A-P$. This means P divides CA in some ratio.
No, let's reconsider. The line passing through C(3,-2) and P(-1,-5) contains the point A.
The distance $CP=5$. The point A is on the circle at a distance of $r=6$ from C.
The point of minimum distance A is on the line segment from P, extending away from C.
The distance $PA = r - d = 6 - 5 = 1$.
The point A divides the segment CP externally, with P between C and A. A is on the line CP. P divides CA in the ratio $5:1$.
Let $A=(x,y)$. Then $P = \left(\frac{1 \cdot C_x + 5 \cdot A_x}{1+5}, \frac{1 \cdot C_y + 5 \cdot A_y}{1+5}\right)$.
$-1 = \frac{1(3)+5x}{6} \implies -6 = 3+5x \implies 5x = -9 \implies x=-9/5$. $-5 = \frac{1(-2)+5y}{6} \implies -30 = -2+5y \implies 5y = -28 \implies y=-28/5$. So, A is $(-9/5, -28/5)$.