Question

. Show that the circles
x^2 + y^2 - 2x - 2y - 7 = 0,
3x^2 + 3y^2 - 8x + 29y = 0 are orthogonal.​

Answer 1

Step-by-step explanation:

Thunderstorm, a violent short-lived weather disturbance that is almost always associated with lightning, thunder, dense clouds, heavy rain or hail, and strong gusty winds. ... Thunderstorms arise when layers of warm, moist air rise in a large, swift updraft to cooler regions of the atmosphere.

Answer 2

$\large\underline{\sf{Solution-}}$

Given circles are

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 2x - 2y - 7 = 0 - - - (1)$

and

$\rm :\longmapsto\: {3x}^{2} + {3y}^{2} - 8x + 29y = 0 - - - (2)$

Consider circle (1),

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 2x - 2y - 7 = 0$

On comparing with

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$

we get,

$\red{\rm :\longmapsto\:g = - 1} \\ \red{\rm :\longmapsto\:f = - 1} \\ \red{\rm :\longmapsto\:c = - 7}$

Consider circle (2)

$\rm :\longmapsto\: {3x}^{2} + {3y}^{2} - 8x + 29y = 0$

Divide whole equation by 3, we get

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - \dfrac{8}{3} x + \dfrac{29}{3} y = 0$

On comparing with

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + 2g_1x + 2y_1y + c_1 = 0$

we get

$\red{\rm :\longmapsto\:g_1 = - \dfrac{8}{6} } \\ \red{\rm :\longmapsto\:f_1 = \dfrac{29}{6} } \\ \red{\rm :\longmapsto\:c_1 = 0}$

Now, to show that two circles are Orthogonal, we have to show that

$\boxed{\red{\bf :\longmapsto\:gg_1 + ff_1 = c + c_1}}$

Now,

Consider

${\red{\bf :\longmapsto\:gg_1 + ff_1 }}$

$\rm \: \: = \: \: ( - 1) \times \dfrac{( - 8)}{3} + ( - 1) \times \dfrac{29}{3}$

$\rm \: \: = \: \: \dfrac{8}{3} - \dfrac{29}{3}$

$\rm \: \: = \: \: - \dfrac{21}{3}$

$\rm \: \: = \: \: - \: 7$

$\bf\implies \:gg_1 + ff_1 = - 7 - - - (3)$

Consider

$\red{\bf :\longmapsto\:c + c_1}$

$\rm \: \: = \: \: - 7 + 0$

$\rm \: \: = \: \: - 7$

$\bf\implies \:c + c_1 = - 7 - - - (4)$

From (3) and (4), we concluded that

$\boxed{\red{\bf \longmapsto\:gg_1 + ff_1 = c + c_1}}$

Hence,

• Two circles intersects each other orthogonally.