Find the equation of a circle with center (4, | vedclass

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(D) Let the circle be $(x - 4)^2 + (y - 3)^2 = r^2$.The center of this circle is $C_1 = (4, 3)$ and its radius is $r$.The given circle is $x^2 + y^2 =

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$x^2 + y^2 - 8x - 6y + 20 = 0$

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(D) Let the circle be $(x - 4)^2 + (y - 3)^2 = r^2$.The center of this circle is $C_1 = (4, 3)$ and its radius is $r$.The given circle is $x^2 + y^2 = 1$,which has center $C_2 = (0, 0)$ and radius $R = 1$.Since the circles touch internally,the distance between their centers must be equal to the difference of their radii:$C_1C_2 = |R - r|$Distance $C_1C_2 = \sqrt{(4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.So,$5 = |1 - r|$.This gives $1 - r = 5$ or $1 - r = -5$.$r = -4$ (not possible) or $r = 6$.Thus,the equation is $(x - 4)^2 + (y - 3)^2 = 6^2$.$x^2 - 8x + 16 + y^2 - 6y + 9 = 36$.$x^2 + y^2 - 8x - 6y + 25 = 36$.$x^2 + y^2 - 8x - 6y - 11 = 0$.

Find the equation of a circle with center $(4, 3)$ that touches the circle $x^2 + y^2 = 1$ internally.

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