find the equation of the circle having centre in the first quadrant touching the x axis having a common tangent y=√3x + 4 with the circle X^2 + Y^2 + 4 x + 4 Y + 4 = 0. such that the distance between the two circles along the x axis is 3 units.
Answers
Step-by-step explanation:
Is that y = (√3)x+4 ?
The other circle is clearly
(x+2)^2 + (y+2)^2 = 4
and the way the question is worded, it appears that the first circle is also tangent to the x-axis at (1,0), making it
(x-1)^2 + (y-k)^2 = k^2
Now it should not be too hard to find k so that the desired line is a common tangent.
Answer:
We haveC:x2+y2+4x+4y+4=0Centre is (−2,2) and radius is (−2)2+(−2)2−4−−−−−−−−−−−−−−√=4+4−4−−−−−−−√=2If y=3–√x+4 touches C then perpendicular from centre to line must beequal to radius of circle.y=3–√x+4 ⇒3–√x−y+4=0Length of perpendicular=∣∣3√×(−2)−2+4∣∣(3√)2+(−1)2√=∣∣23√+2∣∣2=2(3√+1)2=3–√+1 which is not equal to radius of cirlcle. Hence line cannot be tangentto cirlce.Hence information given in question is incorrect.Your question is wrong.We haveC:x2+y2+4x+4y+4=0Centre is -2,2 and radius is -22+-22-4=4+4-4=2If y=3x+4 touches C then perpendicular from centre to line must beequal to radius of circle.y=3x+4 ⇒3x-y+4=0Length of perpendicular=3×-2-2+432+-12=23+22=23+12=3+1 which is not equal to radius of cirlcle. Hence line cannot be tangentto cirlce.Hence information given in question is incorrect.
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