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There are two circles whose equation are `x^2+y^2=9`and `x^2+y^2-8x-6y+n^2=0,n in Zdot`If the two circles have exactly two common tangents, then the number ofpossible values of `n`is2 (b) 8(c) 9 (d)none of theseA. 2B. 8C. 9D. none of these |
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Answer» Correct Answer - C The coordinates of the centres and radii of the circles are: `{:("Centres:",C_(1)(0, 0),C_(2)(4, 3)),("Radii",r_(1)=3,r_(2)=sqrt(25-n^(2))-5 lt n lt 5),(,,):}` Given circles will have exactly two common tangents, if `|r_(1)-r_(2)| lt C_(1) C_(2) lt R_(1)+r_(2)` `rArr |3-sqrt(25-n^(2))| lt 5` is true for all `n in (-5, 5)`. Now, `5 lt 3 + sqrt(25-n^(2)) ` `rArr 2 lt sqrt(25-n^(2))` `rArr 4 lt 25 - n^(2)` `rArr n^(2)-21 lt 0 ` `rArr -sqrt (21) lt n lt sqrt(21) rArr n = pm 4, pm 3, pm 2, pm1, 0`. Hence, n can take 9 integral values. |
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