Name: Understanding Parametric Form of a Circle
Uploaded: 2019-07-19
Duration: 6 min 4 s
Description: This video deals with the concept of parametric equations of circles, how they are written, how we can write the parametric form of a circle from cartesian f...

Question

Show that, four common tangents can be drawn for the circles given by

x^{2} + y^{2} - 14 x + 6 y + 33 = 0

and

x^{2} + y^{2} + 30 x - 2 y + 1 = 0

and find the internal and external centres of similitude.

Accepted Answer

Given, equation of circles are

$x^{2} + y^{2} - 14 x + 6 y + 33 = 0$ and $x^{2} + y^{2} + 30 x - 2 y + 1 = 0$

Let us find the centre and radius of circle $x^{2} + y^{2} - 14 x + 6 y + 33 = 0$

$x^{2} + y^{2} + 2 \times (- 7) x + 2 \times 3 y + 33 = 0$

$\Rightarrow$ Centre of this circle $= C_{1} (- g, - f) = C_{1} (7, - 3)$

And, radius of this circle $= r_{1} = \sqrt{g^{2} + f^{2} - c} = \sqrt{{(- 7)}^{2} + {(3)}^{2} - 33} = \sqrt{49 + 9 - 33} = \sqrt{25} = 5$ units

Now, let us find the centre and radius of circle $x^{2} + y^{2} + 30 x - 2 y + 1 = 0$

We can also write this equation as

$x^{2} + y^{2} + 2 \times (15) x + 2 \times (- 1) y + 1 = 0$

$\Rightarrow$ Centre of this circle $= C_{2} (- 15, 1)$

And, radius of this circle $= r_{2} = \sqrt{{(15)}^{2} + 1 - 1} = 15$ units

For two circles to have common tangents we must have

$C_{1} C_{2} > r_{1} + r_{2}$

Now, $C_{1} C_{2} = \sqrt{{(7 + 15)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{484 + 16} = \sqrt{500}$ units

And, $r_{1} r_{2} = 15 \times 5 = 75$ units

Since $C_{1} C_{2} > r_{1} + r_{2}$

Hence, we can draw four common tangents to the given circles.

The internal similitude is the point where transverse common tangents meet. Let the required point be denoted by $O$ then the point $O$ divided $C_{1} C_{2}$ internally in the ratio of $r_{1} : r_{2}$

i.e., $5 : 15 = 1 : 3$

So, using the section formula,

$O (x, y) = (\frac{1 (- 15) + 3 (7)}{1 + 3}, \frac{1 (1) + 3 (- 3)}{1 + 3}) = (\frac{6}{4}, \frac{- 8}{4}) = (\frac{3}{2}, - 2)$ .

Now, the external centres of similitude divides the line at centres $C_{1} C_{2}$ in the ratio $r_{1} : r_{2}$ externally.

Thus, the required external centres of similitude $= (\frac{r_{1} x_{2} - r_{2} x_{1}}{r_{1} - r_{2}}, \frac{r_{1} y_{2} - r_{2} y_{1}}{r_{1} - r_{2}}) = (\frac{(1) (- 15) - (3) (7)}{- 2}, \frac{(1) (1) - 3 (- 3)}{- 2}) = (18, - 5)$

Show that, four common tangents can be drawn for the circles given by $x^{2} + y^{2} - 14 x + 6 y + 33 = 0$ and $x^{2} + y^{2} + 30 x - 2 y + 1 = 0$ and find the internal and external centres of similitude.

Important Questions on Circle

Learn and Prepare for any exam you want !

Show that, four common tangents can be drawn for the circles given by x2+y2-14x+6y+33=0 and x2+y2+30x-2y+1=0 and find the internal and external centres of similitude.

Important Questions on Circle

Learn and Prepare for any exam you want !

Show that, four common tangents can be drawn for the circles given by $x^{2} + y^{2} - 14 x + 6 y + 33 = 0$ and $x^{2} + y^{2} + 30 x - 2 y + 1 = 0$ and find the internal and external centres of similitude.