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