M(4,3) Figure 2 Figure 2 above shows a sketch of the circles $C_1$ and $C_2$. The circle $C_1$ has the equation $x^2 + y^2 = 4$. The circle $C_2$ has centre M(4, 3) and touches $C_1$ externally, as shown in the figure. (a) Show that the radius of $C_2$ is 3.
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x^2 + y^2 - 4x + 6y - 12 = 0 Rearranging the terms: (x^2 - 4x) + (y^2 + 6y) = 12 To complete the square for x, we need to add (4/2)^2 = 4 to both sides: (x^2 - 4x + 4) + (y^2 + 6y) = 12 + 4 (x - 2)^2 + (y^2 + 6y) = 16 To complete the square for y, we need Show more…
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