what will be the point of contact of the given circles : x2 +y2 -6x -6y +10 =0 and x2+y2=2 - Maths - Conic Sections

Question

what will be the point of contact of the given circles :
x2 +y2 -6x -6y +10 =0 and
x2+y2=2

Rizwan Ansari · Accepted Answer

The given two circles are

$x^{2} + y^{2} - 6 x - 6 y + 10 = 0$ ...... (1)

$x^{2} + y^{2} = 2$ ...... (2)

Now, we will compare the above two circles with the following standard equation

${a x}^{2} + b y^{2} + 2 g x + 2 f y + c = 0$

For first circle,

$g = - 3, f = - 3 and c = 10$

$Centre = (- g, - f) = (3, 3) = (x_{1}, y_{1}) Radius = \sqrt{g^{2} + f^{2} - c} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2} - 10} = 2 \sqrt{2}$

For second circle,

$Centre = (0, 0) = (x_{2}, y_{2}) Radius r_{2} = \sqrt{2}$

Distance between the centre of the two circles is given by

$D = \sqrt{(3 - 0)^{2} + (3 - 0)^{2}} = 3 \sqrt{2}$

$r_{1} + r_{2} = 2 \sqrt{2} + \sqrt{2} = 3 \sqrt{2} = D$

Since $D = r_{1} + r_{2}$ , therefore the circles touch externally and the contact point is given by

$C = (\frac{r_{1} x_{2} + r_{2} x_{1}}{r_{1} + r_{2}}, \frac{r_{1} y_{2} + r_{2} y_{1}}{r_{1} + r_{2}}) = (\frac{2 \sqrt{2} \times 0 + \sqrt{2} \times 3}{2 \sqrt{2} + \sqrt{2}}, \frac{2 \sqrt{2} \times 0 + \sqrt{2} \times 3}{2 \sqrt{2} + \sqrt{2}}) = (1, 1)$

Hence, the contact point is (1,1)