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Simple Co-ordinate geometry problems1.Find the equation of straight line passing through the point P(5,2) with equal intercepts. 1.Method 1Let the equation of straight line be xa+yb=1, a,b≠0(a)If a=b≠0 and the straight line passes through P(5,2),5a+2a=1 and a=7The straight line is x+y=7.(b)If a=b=0, the straight line passes through the origin and the equation is 2x-5y=0 .Method 2Let the equation of straight line passing through the point P(5,2) bey-2=mx-5Since x-intercept = y-intercept-2m+5=2-5m , where m≠05m2-3m-2=0 ∴m=-25 or m=1The equations are y-2=-25x-5 or y-2=1x-52x-5y=0 or 2x-5y=0 .2.Consider a circle with center (4,5) and with radius of 8 If the tangent lines of the circle has slope 13. Find :(a)the points of contact of these tangent lines and the circle,(b)the equation of the tangent lines.2.(a)Method 1The circle is C: x-42+ y-52=64Since the tangent slope is 13 , the slope of the line perpendicular to the tangent is -3.Let this diameter line passes through the centre 4,5 be L. Then L: y-5=-3x-4Solving,C: x-42+ y-52=64 ….(1)L: y-5=-3x-4 ….(2)2↓1,x-42+ 9x-42=6410x-42=64x-4=±810=±4105x=20±4105 and y=25±12105The required points are 20+4105, 25+12105 or 20-4105, 25-12105Method 2The circle is C: x-42+ y-52=64Differentiate with respect to x,2x-4+2y-5dydx=0?dydx=-x-4y-5Since the tangent slope is 13 , -x-4y-5=13We get a straight line L: y=-3x+17 (This line is not the tangent line, but the diameter line passing (4,5) and the point of contact. In fact we just get a line which is good for the given.)Substitute in C, we have x-42+ -3x+17-52=6410 x2-80x+96=0 or 5x2-40 x-48=0 x=20±4105 and y=25±12105The required points are 20+4105, 25+12105 or 20-4105, 25-12105Method 3The circle in parametric form is C: x=4+8cosθy=5+8sinθ, 0≤θ<360°dxdt=-8sinθdydt=8cosθ?dydx=8cosθ-8sinθ=cotθSince the tangent slope is 13 , cotθ=13?tanθ=3?cosθ=±110=±1010sinθ=±310=±31010x=4+8cosθ=4+8±1010y=5+8sinθ=5+8±31010The required points are 20+4105, 25+12105 or 20-4105, 25-12105.Method 4The circle in parametric form is C: x=4+8cosθy=5+8sinθ, 0≤θ<360°Let the pencil of straight lines with slope 13 be L: y=13x+c Hence, we have:5+8sinθ=134+8cosθ+c8cosθ-24sinθ=11-3cWriting in subsidiary angles, 82+242cosθ-α=11-3c where tanα=248=3Since the straight line touches the circle at only one point, we have cosθ-α=1 or-1Therefore, ±82+242=11-3cc=11+8 103(b)Method 1Use point-slope form, in (a) we get slope = 13, Points of contact = 20±4105, 25±12105Hence equations of tangent are:y-25±12105=13x-20±4105y=13x+11±8 103If there is no need to find the points of contact, the following methods are also of interest.Method 2Let the pencil of straight lines with slope 13 be L: y=13x+c ….(1) It touches the circle C: x-42+ y-52=64 ….(2)(1)↓(2),x-42+ 13x+c-52=6410 x29+2 c 3-34 3x+c2-10 c-23=0Since L touches C at one point, ?=2 c 3-34 32-4×109×c2-10 c-23=0-4 c2+88 c3+6923=012 c2-88c-692=0 c=11+8 103 or c=11-8 103Hence the equations of tangent are:y=13x+11±8 103Method 3The circle in parametric form is C: x=4+8cosθy=5+8sinθ, 0≤θ<360°Let the pencil of straight lines with slope 13 be L: y=13x+c Hence, we have:5+8sinθ=134+8cosθ+c8cosθ-24sinθ=11-3cWriting in subsidiary angles, 82+242cosθ-α=11-3c where tanα=248=3For each c we may have no solution, one solution or two solutions for the straight line.Since the straight line touches the circle at only one point, we have cosθ-α=1 or-1.(For cosθ-α=0, the straight line cuts the circle at two points! You get a diameter equation which is parallel to the tangent lines we are interested. Investigate yourselves.))Therefore, ±82+242=11-3cc=11+8 103Hence the equations of tangent are:y=13x+11±8 1033.Find the value(s) of m such that 2m3+m2-mx2+m3-m2+2my2-8m+18=0represents the equation of a circle.3.Ax2+By2+C=0 represents a circle if and only if A=B≠0 and the radius > 0.Now,2m3+m2-m=m3-m2+2mm3+2m2-3m=0m m-1 m+3=0m=0 or m=1 or m=-3(a)When m=0, 2m3+m2-m=m3-m2+2m=0, no equation existed and m=0 is rejected.(b)When m=1, The circle is 2x2+2y2+10=0 or x2+y2+5=0The radius =-5 which is imaginary and m=1 is rejected.(c)When m=-3, The circle is -42x2-42y2+42=0 or x2+y2-1=0The radius is 1 . This satisfies all conditions of the given.4.Find the equation of a circle C1 passing through the intersection points of L: x-3y+4=0C: x2+y2+2x-6y+2=0and with the smallest area.4.Method 1The intersection point of L: x-3y+4=0C: x2+y2+2x-6y+2=0are-3 11-25, 6-115 or 3 11-25, 11+65 (working steps omitted)The smallest circle that can be formed should use these two points as diameter.Using the diameter form, the required circle is:C1: y-6-115x--3 11-25×y-11+65x-3 11-25=-1 C1: y-6-115y-11+65+x--3 11-25x-3 11-25=0C1: x2+y2+4 x5-12 y5-145=0Method 2Let the system of circles passing through L and C beC1: x2+y2+2x-6y+2+λx-3y+4=0C1: x2+y2+2+λx-6+3λy+2+4λ=0(Method 2A) If this circle has the smallest area, the radius r is also smallest.r2=-2+λ22+6+3λ22-2+4λ=5 λ2+12λ+162=52λ2+125λ+165=52λ+652-3625+162=52λ+652+16425Hence r is smallest when λ=-65. (Method 2B) If this circle has the smallest area, the radius r is also smallest.Hence the centre of C1 = -2+λ2,6+3λ2 must be on the line L.We therefore have -2+λ2-36+3λ2+4=0Hence r is smallest when λ=-65. The required circle is: C1: x2+y2+2-65x-6+3-65y+2+4-65=0C1: x2+y2+4 x5-12 y5-145=0Method 3The centre, G, of C = -1,3Let L1 be the line perpendicular to L and passing through G.Gradient of L = 13 and Gradient of L1= -3.Hence, L1:y-3=-3x+1 or y=-3xSolving L and L1, we get A-25,65. This is the centre of the required circle.Let the system of circles passing through L and C beC1: x2+y2+2x-6y+2+λx-3y+4=0C1: x2+y2+2+λx-6+3λy+2+4λ=0The centre is also A-2+λ2,6+3λ2=-25,65-2+λ2=-25 and λ=-65. The required circle is: C1: x2+y2+2-65x-6+3-65y+2+4-65=0C1: x2+y2+4 x5-12 y5-145=0 ................
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