Equation of family of circles passing through origin
Answers
Answer:
We will learn how to form the equation of a circle passes through the origin.
The equation of a circle with centre at (h, k) and radius equal to a, is (x - h)2
2
+ (y - k)2
2
= a2
2
.
When the centre of the circle coincides with the origin i.e., a2
2
= h2
2
+ k2
2
Let O be the origin and C(h, k) be the centre of the circle. Draw CM perpendicular to OX.
Circle Passes through the Origin
In triangle OCM, OC 2
2
= OM 2
2
+ CM 2
2
i.e., a 2
2
= h 2
2
+ k 2
2
.
Therefore, the equation of the circle (x - h)2
2
+ (y - k)2
2
= a2
2
becomes
(x - h) 2
2
+ (y - k) 2
2
= h 2
2
+ k 2
2
⇒ x 2
2
+ y 2
2
- 2hx – 2ky = 0
The equation of a circle passing through the origin is
x 2
2
+ y 2
2
+ 2gx + 2fy = 0 ……………. (1)
or, (x - h)2
2
+ (y - k)2
2
= h2
2
+ k2
2
…………………………. (2)
We clearly see that the equations (1) and (2) are satisfied by (0, 0).