Question

the intercept made by circle with centre (2,3) and radius 6 on Y-axis is?

Answer 1

Answer:

the coordinates of the intercept are ( 0,  3+4$\sqrt{2}$ )  or ( 0, 3-4$\sqrt{2}$ )

Step-by-step explanation:

answer is in attachment

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Answer 2

Answer:

(0, 3+$4\sqrt{2}$) and (0, 3 - $4\sqrt{2}$)

Step-by-step explanation:

Given:

center of the circle (2,3)

radius of circle = 6

To find: intercept made by a circle on the y-axis

Solution:

we know,

equation of the circle when center and radius are given = $(x-a)^{2} - (y-b)^{2} = r^{2}$

putting the values in the above equation

we get,

$(x - 2)^{2} + (y - 3)^{2} = 6^{2}$

=> $x^{2} - 4x^{} + 4 + y^{2} - 6y^{} + 9 = 36$

=> $x^{2} + y^{2} - 4x^{} - 6y^{} - 23 = 0$

Now, we have to find the y-intercept of the circle.

∴ put x = 0 in the equation (i)

we get,

$y^{2} - 6y^{} - 23 = 0$

=> y = $\frac{-b + \sqrt{b^{2} - 4ac} }{2a}$ and $\frac{-b - \sqrt{b^{2} - 4ac } }{2a}$

were, a = 1, b = -6, c = -23

putting the values

we get,

y = $\frac{6 + \sqrt{128} }{2}$ and $\frac{6 - \sqrt{128} }{2}$

=> y = $\frac{2 (3 +4 \sqrt{2} )}{2}$ and $\frac{2 (3 - 4\sqrt{2}) }{2}$

Hence, Circle intercept the Y-axis at (0, 3 + $4\sqrt{2}$) and (0, 3 - $4\sqrt{2}$). Ans

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