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प्रश्न
The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is ______.
रिकाम्या जागा भरा
उत्तर
The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is `(x - 3)^2 + (y + 4)^2 = (45/13)^2`.
Explanation:
Given equation of the line is 5x + 12y – 12 = 0 and the centre is (3, – 4)
CP = radius of the circle
`|(5 xx 3 + 12 xx (-4) - 12)/sqrt((5)^2 + (12)^2)|` = r
⇒ `|(15 - 48 - 12)/13|` = r
⇒ `|(-45)/13|` = r
⇒ `r^2 = 2025/169`
So, the equation of the circle is `(x - 3)^2 + (y + 4)^2 = (45/13)^2`
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