The equation of the circle centered at (4,3) | vedclass

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Question

(A) Given the equation of the circle $x^{2}+y^{2}=1$,the center is $O(0,0)$ and the radius $r_{1} = 1$.Let the center of the required circle be $C(4,3

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation of the circle $x^{2}+y^{2}=1$,the center is $O(0,0)$ and the radius $r_{1} = 1$.Let the center of the required circle be $C(4,3)$ and its radius be $r_{2}$.The distance between the centers $O$ and $C$ is $OC = \sqrt{(4-0)^{2} + (3-0)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$.Since the circles touch externally,the distance between their centers is equal to the sum of their radii:$OC = r_{1} + r_{2}$$5 = 1 + r_{2}$$r_{2} = 4$.The equation of the circle with center $(h,k) = (4,3)$ and radius $r = 4$ is:$(x-4)^{2} + (y-3)^{2} = 4^{2}$$x^{2} - 8x + 16 + y^{2} - 6y + 9 = 16$$x^{2} + y^{2} - 8x - 6y + 9 = 0$.

The equation of the circle centered at $(4,3)$ touching the circle $x^{2}+y^{2}=1$ externally is:

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