The lengths that intercepts made by the circle x(^2) + y(^2) + 2gx + 2fy + c = 0 through X and also Y axes are 2(mathrmsqrtg^2 - c) and also 2(mathrmsqrtf^2 - c) respectively.

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Proof:

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

Clearly, the centre of the one is c (-g, -f) and also the radius = (mathrmsqrtg^2 + f^2- c)

Let abdominal muscle be the intercept do by the offered circle ~ above x-axe.Since ~ above x-axis, y = 0. Therefore, x-coordinates the the point out A and B space theroots of the equation x(^2) + 2gx + c = 0. Intercepts on the Axes make by a Circle

Let x(_1) and also x(_2) be the x-coordinates of the points A and also Brespectively. Then, x(_1) and x(_2) likewise the root of the equation x(^2) + 2gx + c = 0.

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

Clearly the intercept top top x-axis = abdominal muscle

= x(_2) - x(_1) = (mathrmsqrt(x_2 - x_1)^2)

= (mathrmsqrt(x_2 +x_1)^2 - 4x_1x_2)

= (mathrmsqrt4g^2 - 4c)

= 2(mathrmsqrtg^2 - c)

Therefore, the intercept make by the one (1) top top thex-axis = 2(mathrmsqrtg^2 - c)

Again,

Let DE be the intercept made by the given circle on y-axe.Since on y-axis, x = 0. Therefore, y-coordinates the the points D and also E room theroots the the equation y(^2) + 2fy + c = 0.

Let y(_1) and y(_2) be the x-coordinates the the point out D and also Erespectively. Then, y(_1) and also y(_2) likewise 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 on y-axis = DE

= y(_2) - y(_1) = (mathrmsqrt(y_2 - y_1)^2)

= (mathrmsqrt(y_2 + y_1)^2 – 4y_1y_2)

= (mathrmsqrt4f^2 -4c)

= 2(mathrmsqrtf^2 - c)

Therefore, the intercept made by the circle (1) ~ above the y-axis= 2(mathrmsqrtf^2 - c)

Solved instances to uncover the intercepts make by a provided circle ~ above the co-ordinate axes:

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

Solution:

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

Now comparing the provided equation through the general equation of the circle x(^2) + y(^2) + 2gx + 2fy + c = 0, we obtain g = -2 and f = -3 and also c = -5

Therefore, length of the x-intercept = 2(mathrmsqrtg^2 - c) = 2(mathrmsqrt4 - (-5) ) = 2√9 = 6.

The length of the y-intercept = 2(mathrmsqrtf^2 - c) = 2(mathrmsqrt9 - (-5) ) = 2√14.

2. Find the equation that a circle which touch the y-axis at a distance -3 indigenous the origin and cuts one intercept that 8 units v the optimistic direction the x-axis.

Solution:

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

According come the problem, the equation (i) touch the y-axis

Therefore, c = f(^2) ………………… (ii)

Again, the suggest (0, -3) lies ~ above the one (i).

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

9 - 6f + c = 0 …………………… (iii)

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

Now putting f = 3 in (i) us get, c = 9

Again, follow to the trouble the equation of the circle (i) cut an intercept the 8 units v the positive direction that x-axis.

Therefore,

2(mathrmsqrtg^2 - c) = 8

⇒ 2(mathrmsqrtg^2 - 9) = 8

⇒ (mathrmsqrtg^2 - 9) = 4

⇒ g(^2) - 9 = 16,

⇒ g(^2) = 16 + 9

⇒ g(^2) = 25

⇒ g = ±5.

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Hence, the compelled equation the the one is x^2 + y^2 ± 10x + 6y + 9 = 0.

The Circle

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