Question

Equation of the circle centered at $(4, 3)$ touching the circle $x^2+y^2-1$ externally, is _____

• A. $x^2 + y^2 + 8x - 6y + 9 = 0$
• B. $x^2+y^2-8x + 6y + 9 = 0$
• C. $x^2 + y^2 - 8t - 6y + 9 = 0$
• D. $x^2 + y^2 + 8x + 6y + 9 = 0$

Answer 1

Answer: (C)

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

Answer 2

Given that, equation of circle
$x^{2}+y^{2}=1$

Centre at $O \rightarrow(0,0)$
Radius $=O A=1$
Also, the centre of another circle $\rightarrow C(4,3)$ both circle touch externally. Then, distance between centres $=O C$.
$=\sqrt{(4-0)^{2}+(3-0)^{2}}=\sqrt{16-9}=5$
Now, $A C=O C-O A$
$A C=5-1=4$
So, the radius of other circle is 4 .
Now, the equation of other circle touch externally to the circle $x^{2}+y^{2}=1$ is,
$(x-4)^{2}+(y-3)^{2}=16$
$\Rightarrow x^{2}+y^{2}-8 x-6 y+9=0$