Question

Find the locus of the middle points of the chords of the circle $x^2+y^2+2gx+2fy+c=0$ which subtend a right angle at the origin.

Answer 1

Answer:

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Step-by-step explanation:

Solution

Let AB be the chord of the circle and P be the midpoint of AB.

It is known that perpendicular from the center bisects a chord.

Thus △ACP is a right-angled triangle.

Now AC=BC= radius.

The equation of the give circle can be written as

(x−1)2+(y−2)2=16

Hence, centre C=(1,2) and radius =r=4 units.

PC=ACsin600

=rsin600

=4(23)

=23 units

Therefore, PC=23

⇒PC2=12

⇒(x−1)2+(y−2)2=12

⇒x2+y2−2x−4y+5=12

⇒x2+y2−2x−4y−7=0