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