Question

Find the equation of a circle with centre (4,3) touching the circle x² + y² = 1


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Answer 1

$\bf \blue{hello \: dear}$

$\bf \pink{solution \: is \: in \: this \: given \: picture}$

$\boxed{hope \: it \: helps \: you}$

$\underline{rajukumar}$

Answer 2

Step-by-step explanation:

x^2 + y^2 = 1 is the equation of the circle with centre (0,0) and radius 1 unit.

If two circles touch each each other then the point of contact and the centres of the circles lie on the same line.

So distance between the centres is equal to the sum of their radii.

So distance between (0,0) and (4,3) is,

=> √{(4–0)^2+(3–0)^2}

=> √(16+9)

=> 5

So radius of the required circle is 5 - 1 = 4

So equation of the circle with centre (4,3) and radius 4 units is

(x-4)^2 + (y-3)^2 = 4^2

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

Hope it helps!