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Find the equation of circles that touch both the axes and pass through (− 4, −2) in general form - Mathematics

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प्रश्न

Find the equation of circles that touch both the axes and pass through (− 4, −2) in general form

बेरीज

उत्तर


Since the circle touches both the axes

Its centre will be (r, r) and the radius will be r.

Here centre = C = (r, r) and point on the circle is A = (− 4, −2)

CA = r

⇒ CA2 = r2

(i.e) (r + 4)2 + (r + 2)2 = r2

⇒ r2 + 8r +16 + r2 + 4r + 4 – r2 = 0

(i.e) r2 + 12r + 20 = 0

(r + 2)(r + 10) = 0

⇒ r = − 2 or − 10

When r = − 2, the equation of the circle will be (x + 2)2 + (y + 2)2 = 22

(i.e) x2 + y2 + 4x + 4y + 4 = 0

When r = −10, the equation of the circle will be (x + 10)2 + (y + 10)2 = 102

(i.e) x2 + y2 + 20x + 20y + 100 = 0

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पाठ 5: Two Dimensional Analytical Geometry-II - Exercise 5.1 [पृष्ठ १८२]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 5 Two Dimensional Analytical Geometry-II
Exercise 5.1 | Q 3 | पृष्ठ १८२

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