Question

A circle touching the x-axis at $(3, 0)$ and making an intercept of length $8$ on the y-axis passes through the point :

• A. (3, 10)
• B. (2,3)
• C. (1,5)
• D. (3,5)

Answer 1

The Correct Option is A

Equaiton of circles are

Answer 2

We can see that our circle touches the x-axis at (3,0) and the circle makes an intercept of length 8 on the y-axis.

 

 

 Let us notice the figure which is drawn based on the above information.

Let's assume the radius of the circle is r. As the circle meet the x-axis, it is tangent to the circle and so a line drawn from the centre of the circle to (3,0) is perpendicular to the x-axis. As the radius is as r, our centre must be O(3,r)

Now let us draw a perpendicular from the centre to the y-axis. As a perpendicular line from the centre to a secant to the circle bisects the secant, point P is the midpoint of the intercept made by the circle on the y-axis.

Now, consider the triangle ΔOPQ. As it is a right-angled triangle, we can apply Pythagoras' theorem to this triangle.

Let us consider the Pythagoras theorem,

A sum of squares of sides is equal to the square of the hypotenuse, that is 

a²+b²=c²

Using the above formula, we get

⇒ OP²+PQ²=OQ²

⇒ 3²+4²=r²

⇒ 9+16=r²

⇒ r² = 25

⇒ r = 5

So, the radius of the circle  5 units.

As we got the radius r=5 units, we can see that the point C becomes

⇒ (3,2r) = (3,2×5) = (3,10)

Which is the same as the point in Option A.

 

So, the correct answer is “Option A”.