The curve $C$ has equation $\dfrac{(x-2)^2}{2} + (y-3)^2 = 1$. Sketch $C$, [3] Hence state the range

Question

The curve $C$ has equation $\dfrac{(x-2)^2}{2} + (y-3)^2 = 1$. Sketch $C$, [3] Hence state the range of positive values of $k$ such that $(x-2)^2 + (y+3)^2 = k^2$ intersects $C$ at twice. [2]

Accepted Answer

(ii) $5 < k < 7$

Worked Example 2

Worked Example 2

Video Solution:

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