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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Find the equation of the circle with centre at (3,1) and touching the line 8x − 15y + 25 = 0

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प्रश्न

Find the equation of the circle with centre at (3,1) and touching the line 8x − 15y + 25 = 0

बेरीज
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उत्तर

The centre of the circle is (3, 1) and it is touching the line 8x − 15y + 25 = 0.

∴ radius = perpendicular distance from (3, 1) to the line 8x − 15y + 25 = 0

= `|(8(3) + (-15)(1) + 25)/sqrt(8^2 + (-15)^2)|`

= `|(24 - 15 + 25)/sqrt(64 + 225)|`

= `|34/17|`

= 2

∴ the equation of the circle is

(x − 3)2 + (y − 1)2 = 22

∴ x2 − 6x + 9 + y2 − 2y + 1 = 4

∴ x2 + y2 − 6x − 2y + 6 = 0

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पाठ 6: Circle - Exercise 6.1 [पृष्ठ १२९]

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