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Question: The locus of the center of the circles such that the point \[\left( {2,3} \right)\] is the midpoint ...

The locus of the center of the circles such that the point (2,3)\left( {2,3} \right) is the midpoint of the chord 5x+2y=165x + 2y = 16 is:
A. 2x−5y−11=02x - 5y - 11 = 0
B. 2x+5y−11=02x + 5y - 11 = 0
C. 2x+5y+11=02x + 5y + 11 = 0
D. None

Explanation

Solution

First of all, find the slope of the given chord and then find the slope of the line joining the centre of the circle and the midpoint of the chord. Use the condition of the perpendicularity of two lines to find the required locus as these two lines are perpendicular to each other.

Complete step-by-step answer:

The given equation of the chord is 5x+2y=165x + 2y = 16 which can be written as 5x+2y−16=05x + 2y - 16 = 0.

The slope of the line ax+by+c=0ax + by + c = 0 is−ab\dfrac{{ - a}}{b}.

So, the slope of the chord is −52\dfrac{{ - 5}}{2}

Let the centre of the circle be (−g,−f)\left( { - g, - f} \right) as shown in the below figure:

The midpoint of the chord is (2,3)\left( {2,3} \right).

Now, consider the slope of the line joining the midpoint on the chord and the centre of the circle (−g,−f)\left( { - g, - f} \right) is 3−(−f)2−(−g)=3+f2+g\dfrac{{3 - \left( { - f} \right)}}{{2 - \left( { - g} \right)}} = \dfrac{{3 + f}}{{2 + g}}

The line joining the midpoint of the chord and the centre of the circle and the chord are perpendicular to each other.

So, the product of their slopes is equal to −1 - 1 i.e.,

⇒−52×3+f2+g=−1 \Rightarrow \dfrac{{ - 5}}{2} \times \dfrac{{3 + f}}{{2 + g}} = - 1

⇒−5(3+f)=−2(2+g)\Rightarrow - 5\left( {3 + f} \right) = - 2\left( {2 + g} \right)

⇒15+5f=4+2g\Rightarrow 15 + 5f = 4 + 2g

⇒2g−5f+4−15=0\Rightarrow 2g - 5f + 4 - 15 = 0

∴2g−5f−11=0\therefore 2g - 5f - 11 = 0

Now, to find the locus of all the circle put values g=xg = x and f=yf = y we have

2x−5y−11=02x - 5y - 11 = 0

Hence the locus of the centre of the circles such that the point (2,3)\left( {2,3} \right) is the midpoint of the chord 5x+2y=165x + 2y = 16 is 2x−5y−11=02x - 5y - 11 = 0.

Thus, the correct option is A. 2x−5y−11=02x - 5y - 11 = 0

Note: If two lines are perpendicular then the product of the slopes of the two lines are equal to −1 - 1 which is called the condition of perpendicularity of two lines. The slope of the line ax+by+c=0ax + by + c = 0 is−ab\dfrac{{ - a}}{b}.