The lengths the intercepts made by the one x$$^2$$ + y$$^2$$ + 2gx + 2fy + c = 0 v X and Y axes room 2$$\mathrm\sqrtg^2 - c$$ and 2$$\mathrm\sqrtf^2 - c$$ respectively.

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Proof:

Let the offered equation the the circle it is in x$$^2$$ + y$$^2$$ + 2gx + 2fy + c = 0 ………. (1)

Clearly, the center of the circle is c (-g, -f) and the radius = $$\mathrm\sqrtg^2 + f^2- c$$

Let abdominal muscle be the intercept do by the provided circle ~ above x-axe.Since top top x-axis, y = 0. Therefore, x-coordinates of the point out A and also B space theroots of the equation x$$^2$$ + 2gx + c = 0.

Intercepts ~ above the Axes do by a Circle

Let x$$_1$$ and also x$$_2$$ it is in the x-coordinates the the points A and Brespectively. Then, x$$_1$$ and x$$_2$$ also the roots of the equation x$$^2$$ + 2gx + c = 0.

Therefore, x$$_1$$ + x$$_2$$ = - 2g and also x$$_1$$x$$_2$$ = c

Clearly the intercept ~ above x-axis = ab

= x$$_2$$ - x$$_1$$ = $$\mathrm\sqrt(x_2 - x_1)^2$$

= $$\mathrm\sqrt(x_2 +x_1)^2 - 4x_1x_2$$

= $$\mathrm\sqrt4g^2 - 4c$$

= 2$$\mathrm\sqrtg^2 - c$$

Therefore, the intercept make by the one (1) on thex-axis = 2$$\mathrm\sqrtg^2 - c$$

Again,

Let DE it is in the intercept make by the offered circle top top y-axe.Since on y-axis, x = 0. Therefore, y-coordinates that the point out D and E space theroots the the equation y$$^2$$ + 2fy + c = 0.

Let y$$_1$$ and y$$_2$$ be the x-coordinates of the point out D and Erespectively. Then, y$$_1$$ and also y$$_2$$ also the roots of the equation y$$^2$$ + 2fy + c = 0

Therefore, y$$_1$$ + y$$_2$$ = - 2f and y$$_1$$y$$_2$$ = c

Clearly the intercept ~ above y-axis = DE

= y$$_2$$ - y$$_1$$ = $$\mathrm\sqrt(y_2 - y_1)^2$$

= $$\mathrm\sqrt(y_2 + y_1)^2 – 4y_1y_2$$

= $$\mathrm\sqrt4f^2 -4c$$

= 2$$\mathrm\sqrtf^2 - c$$

Therefore, the intercept made by the circle (1) top top the y-axis= 2$$\mathrm\sqrtf^2 - c$$

Solved examples to uncover the intercepts make by a given circle on the co-ordinate axes:

1. Find the size of the x-intercept and y-intercept made by the circle x$$^2$$ + y$$^2$$ - 4x -6y - 5 = 0 through the co-ordinate axes.

Solution:

Given equation of the circle is x$$^2$$ + y$$^2$$ - 4x -6y - 5 = 0.

Now comparing the offered equation through the basic equation that the one x$$^2$$ + y$$^2$$ + 2gx + 2fy + c = 0, we acquire g = -2 and f = -3 and c = -5

Therefore, size of the x-intercept = 2$$\mathrm\sqrtg^2 - c$$ = 2$$\mathrm\sqrt4 - (-5)$$ = 2√9 = 6.

The length of the y-intercept = 2$$\mathrm\sqrtf^2 - c$$ = 2$$\mathrm\sqrt9 - (-5)$$ = 2√14.

2. Find the equation that a circle which touches the y-axis at a distance -3 from the origin and cuts an intercept of 8 units with the hopeful direction of x-axis.

Solution:

Let the equation that the circle be x$$^2$$ + y$$^2$$ + 2gx + 2fy + c = 0 …………….. (i)

According to the problem, the equation (i) touches the y-axis

Therefore, c = f$$^2$$ ………………… (ii)

Again, the allude (0, -3) lies ~ above the circle (i).

Therefore, placing the value of x = 0 and y = -3 in (i) us get,

9 - 6f + c = 0 …………………… (iii)

From (ii) and also (iii), we obtain 9 - 6f + f$$^2$$ = 0 ⇒ (f - 3)$$^2$$ = 0 ⇒ f - 3 = 0 ⇒ f = 3

Now placing f = 3 in (i) we get, c = 9

Again, follow to the difficulty the equation the the circle (i) cut an intercept of 8 units through the hopeful direction that x-axis.

Therefore,

2$$\mathrm\sqrtg^2 - c$$ = 8

⇒ 2$$\mathrm\sqrtg^2 - 9$$ = 8

⇒ $$\mathrm\sqrtg^2 - 9$$ = 4

⇒ g$$^2$$ - 9 = 16,

⇒ g$$^2$$ = 16 + 9

⇒ g$$^2$$ = 25

Hence, the forced equation that the circle is x^2 + y^2 ± 10x + 6y + 9 = 0.

The Circle

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