Properties of Circle Orthogonal to Two Given Circles

To determine the correct statements regarding the circle $S$, let the general equation of the circle $S$ be represented as: $x^2 + y^2 + 2gx + 2fy + c = 0$

The given circles are:

1. $S_1: (x - 1)^2 + y^2 = 16 \implies x^2 + y^2 - 2x - 15 = 0$
2. $S_2: x^2 + y^2 = 1 \implies x^2 + y^2 - 1 = 0$

Step 1: Use the Condition of Orthogonality

Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ intersect orthogonally if: $2g_1g_2 + 2f_1f_2 = c_1 + c_2$

For circle $S$ and $S_2$: Comparing $S_2$ with the general equation, we have $g_2 = 0, f_2 = 0,$ and $c_2 = -1$. Applying the orthogonality condition: $2g(0) + 2f(0) = c - 1 \implies c = 1$

For circle $S$ and $S_1$: Comparing $S_1$ with the general equation, we have $g_1 = -1, f_1 = 0,$ and $c_1 = -15$. Applying the orthogonality condition: $2g(-1) + 2f(0) = c - 15 \implies -2g = c - 15$

Substitute $c = 1$ into the above equation: $-2g = 1 - 15 = -14 \implies g = 7$

Step 2: Use the Given Point to Find $f$

Circle $S$ passes through the point $(0, 1)$. Substituting $(x, y) = (0, 1)$ into the equation of $S$: $0^2 + 1^2 + 2g(0) + 2f(1) + c = 0$ $1 + 2f + c = 0$

Substitute $c = 1$: $1 + 2f + 1 = 0 \implies 2f = -2 \implies f = -1$

Step 3: Find the Centre and Radius of Circle $S$

• The centre of the circle $S$ is given by $(-g, -f)$: $\text{Centre} = (-7, 1)$

• The radius $r$ of the circle $S$ is given by: $r = \sqrt{g^2 + f^2 - c} = \sqrt{7^2 + (-1)^2 - 1} = \sqrt{49 + 1 - 1} = \sqrt{49} = 7$

Conclusion

• The radius of $S$ is $7$.
• The centre of $S$ is $(-7, 1)$.

Therefore, the correct options are (B) and (C).