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.


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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

⇒ g = ±5.

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

 The Circle

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