Question

find the equation of the circle which touches the axis of the X at a distance of 4 from the origin and cut off an intercept of 6 from the axis of y

Answer 1

Refer to the attachment for the answer. In the attachment it is clear that the line from centre always bisects the chord. Therefore length of AB = 6/2 = 3 cm.

Now, Radius of the Circle (OB) = √(AB² + OA²)

= √(3)² + (4)² = 5 unit.

Now, Radius of the circle is 5 unit.

Now, we have the co-ordinates of the centre in which one part is unknown.

Using the equation of circle.

(x - 4)² + (y - a) = radius²

∴ (x - 4)² + (y - a) = 5²

Now, Any point lying on this equation will satisfy its equation.

Therefore,

(4,0) a point on x-axis will satisfy its equation, since it is touching the x - axis at that point.

∴ (4 - 4)² + (0-a)² = 5²

∴ a² = 5²

∴ a = 5

Hence, the equation of the circle will be,

(x - 4)² + (y - 5)² = 25

     

Hope it helps.