The equation of a circle with centre (-4, 3) | vedclass

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Question

(A) The centre of the given circle is $C_1 = (-4, 3)$ and its radius is $r_1$. The circle $x^2 + y^2 = 1$ has centre $C_2 = (0, 0)$ and radius $r_2 =

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$x^2 + y^2 + 8x - 6y + 9 = 0$

Vedclass · Answer

(A) The centre of the given circle is $C_1 = (-4, 3)$ and its radius is $r_1$. The circle $x^2 + y^2 = 1$ has centre $C_2 = (0, 0)$ and radius $r_2 = 1$.The distance between the centres is $d = \sqrt{(-4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = 5$.If the circles touch internally,$r_1 = d + r_2 = 5 + 1 = 6$ or $r_1 = |d - r_2| = |5 - 1| = 4$.Case $1$: If $r_1 = 4$,the equation is $(x + 4)^2 + (y - 3)^2 = 4^2 \implies x^2 + 8x + 16 + y^2 - 6y + 9 = 16 \implies x^2 + y^2 + 8x - 6y + 9 = 0$.Case $2$: If $r_1 = 6$,the equation is $(x + 4)^2 + (y - 3)^2 = 6^2 \implies x^2 + 8x + 16 + y^2 - 6y + 9 = 36 \implies x^2 + y^2 + 8x - 6y - 11 = 0$.Comparing with the options,option $A$ matches the result from Case $1$.

The equation of a circle with centre $(-4, 3)$ and touching the circle $x^2 + y^2 = 1$ is

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