Question

Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y) passing through origin and touching the circle $C$ externally, then the radius of$ T $is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{\sqrt{2}}$

Answer 1

The Correct Option is B

To find the radius of the circle \( T \) centered at \( (0, y) \) that passes through the origin and touches circle \( C \) externally, let's follow these steps:

First, establish the equation of circle \( C \) which is centered at \( (1, 1) \) with radius 1. The general equation of a circle is:

\((x - h)^2 + (y - k)^2 = r^2\)

For circle \( C \), this becomes:

\((x - 1)^2 + (y - 1)^2 = 1^2\)

The circle \( T \) is centered at \( (0, y) \) and passes through the origin \((0, 0)\). The radius of circle \( T \) is therefore:

\(r_T = \sqrt{(0 - 0)^2 + (y - 0)^2} = y\)

The equation of circle \( T \) is:

\(x^2 + (y - k)^2 = y^2\)

Since circle \( T \) is externally tangent to circle \( C \), the distance between the centers of the circles plus the radius of \( T \) should equal the sum of their radii. Hence, we have:

\(\sqrt{(1 - 0)^2 + (1 - y)^2} = 1 + y\)

This simplifies to:

\(\sqrt{1 + (1 - y)^2} = 1 + y\)

Squaring both sides to eliminate the square root, we get:

\(1 + (1 - y)^2 = (1 + y)^2\)

Expanding both sides:

\(1 + 1 - 2y + y^2 = 1 + 2y + y^2\)

This simplifies to:

\(2 - 2y = 1 + 2y\)

Solving for \( y \), rearrange terms:

\(4y = 1\)

Therefore, \(y = \frac{1}{4}\)

The radius of the circle \( T \) is thus \( \frac{1}{4} \).

The correct answer is: \(\frac{1}{4}\).