Question

The centre of the circle given by $\mathbf{r}.(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 15$ and $|\mathbf{r} - (\mathbf{j} + 2\mathbf{k})| = 4$is

• A. (0, 1, 2)
• B. (1, 3, 4)
• C. (–1, 3, 4)
• D. None of these

Answer 1

Answer: (B)

(1, 3, 4)

Answer 2

The equation of a line through the centre $\mathbf{j} + 2\mathbf{k}$ and normal to the given plane is

$\mathbf{r} = \mathbf{j} + 2\mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ .....(i)

This meets the plane at a point for which we must have $((\mathbf{j} + 2\mathbf{k}) + \lambda(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})).(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 15$

⇒ $6 + \lambda(9) = 15 \Rightarrow \lambda = 1$.

Putting $\lambda = 1$ in (i), we obtain the position vector of the centre as $\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$. Hence, the coordinates of the centre of the circle are (1, 3, 4).