{"id":35510,"date":"2022-09-19T17:16:22","date_gmt":"2022-09-19T11:46:22","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35510"},"modified":"2022-09-19T17:16:22","modified_gmt":"2022-09-19T11:46:22","slug":"inter-2nd-year-maths-2b-circle-solutions-ex-1e","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-circle-solutions-ex-1e\/","title":{"rendered":"Inter 2nd Year Maths 2B Circle Solutions Ex 1(e)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Circle Solutions Exercise 1(e) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(e)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nDiscuss the relative position of the following pair of circles.
\ni) x\u00b2 + y\u00b2 – 4x – 6y – 12 = 0
\nx\u00b2 + y\u00b2 + 6x + 18y + 26 = 0.
\nSolution:
\nCentres of the circles are A (2, 3), B(-3, -9)
\nradii are r1<\/sub> = \\(\\sqrt{4+9+12}\\) = 5
\nr2<\/sub> = \\(\\sqrt{9+81-26}\\) = 8
\nAB = \\(\\sqrt{(2+3)^{2}+(3+9)^{2}}\\)
\n= \\(\\sqrt{25+144}\\) = 13 = r1<\/sub> + r2<\/sub>
\n\u2234 The circles touch externally.<\/p>\n

ii) x\u00b2 + y\u00b2 + 6x + 6y + 14 = 0,
\nx\u00b2 + y\u00b2 – 2x – 4y – 4 = 0.
\nSolution:
\nCentres are A (-3, -3), B (1, 2)
\nr1<\/sub> = \\(\\sqrt{9+9-14}\\) = 2,
\nr2<\/sub> = \\(\\sqrt{1+4+4}\\) = 3
\nAB = \\(\\sqrt{(-3-1)^{2}+(-3-2)^{2}}\\)
\n= \\(\\sqrt{16+25}=\\sqrt{41}\\) > r1<\/sub> + r2<\/sub>
\n\u2234 Each circle lies exterior to the other circle.<\/p>\n

iii) (x – 2)\u00b2 + (y + 1)\u00b2 = 9, (x + 1)\u00b2 + (y – 3)\u00b2= 4
\nSolution:
\nCentres are A( 2, -1), B(-1, 3)
\nr1<\/sub> = \\(\\sqrt{4+1+4}\\) = 3, r2<\/sub> = \\(\\sqrt{1+9-6}\\) = 2
\nAB = \\(\\sqrt{(2+1)^{2}+(-1-3)^{2}}\\)
\n= \\(\\sqrt{9+16}\\)
\n= 5 = r1<\/sub> + r2<\/sub>
\n\u2234 The circles touch each other externally.<\/p>\n

iv) x\u00b2 + y\u00b2 – 2x + 4y – 4 = 0, x\u00b2 + y\u00b2 + 4x – 6y – 3 = 0
\nSolution:
\nCentres are A (1, -2), B (-2, 3)
\nr1<\/sub> = \\(\\sqrt{1+4+4}\\) = 3, r2<\/sub> = \\(\\sqrt{4+9+3}\\) = 4
\nAB = \\(\\sqrt{(1+2)^{2}+(-2-3)^{2}}=\\sqrt{9+16}\\)
\n= \\(\\sqrt{9+25}=\\sqrt{34}\\) < r1<\/sub> + r2<\/sub>
\nr1<\/sub> – r2<\/sub> < AB < r2<\/sub> + r1<\/sub>
\n\u2234 The circles intersect each other.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the number of possible common tangents that exist for the following pairs of circles.
\ni) x\u00b2 + y\u00b2 + 6x + 6y + 14 = 0,
\nx\u00b2 + y\u00b2 – 2x – 4y – 4 = 0
\nSolution:
\nc1<\/sub> ( -3, -3) c2<\/sub> = (1, 2)
\n\"Inter
\nThe circles are externally No. of common tangents = 4<\/p>\n

ii) x\u00b2 + y\u00b2 – 4x – 2y + 1 = 0;
\nx\u00b2 + y\u00b2 – 6x – 4y + 4 = 0.
\nSolution:
\nC1<\/sub> (2, 1), C2<\/sub> = (3, 2)
\nr1<\/sub> = \\(\\sqrt{4+1-1}\\) = 2
\nr2<\/sub> = \\(\\sqrt{9+4-4}\\) =3
\n\"Inter
\nC1<\/sub> C2<\/sub> = \\(\\sqrt{(2-3)^{2}+(1-2)^{2}}=\\sqrt{2}\\)
\nC1<\/sub> C2<\/sub> < r1<\/sub> + r2<\/sub> intersect each other 2 tangents (direct)<\/p>\n

iii) x\u00b2 + y\u00b2 – 4x + 2y – 4 = 0;
\nx\u00b2 + y\u00b2 + 2x – 6y + 6 = 0.
\nSolution:
\nC1<\/sub> (2, -1) C2<\/sub> = (-1, 3)
\n\"Inter
\nC1<\/sub> C2<\/sub> = r1<\/sub> + r2<\/sub> touch each other externally; No. of common tangents = 3.<\/p>\n

iv) x\u00b2 + y\u00b2 = 4; x\u00b2 + y\u00b2 – 6x – 8y + 16 = 0
\nSolution:
\n\"Inter
\nTwo circles touch each other.
\n2 – direct tangents
\n1 – transverse tangent (at common point)<\/p>\n

v) x\u00b2 + y\u00b2 + 4x – 6y – 3 = 0
\nx\u00b2 + y\u00b2 + 4x – 2y + 4 = 0.
\nSolution:
\nC1<\/sub> (-2, 3) C2<\/sub> = (-2, 1)
\nr1<\/sub> = \\(\\sqrt{4+9+3}\\) = 4 r2<\/sub> = \\(\\sqrt{4+1-4}\\) = 1
\nC1<\/sub> C2<\/sub> = \\(\\sqrt{(-2+2)^{2}+(3-1)^{2}}\\)
\nC1<\/sub> C2<\/sub> = 2 < 3 = r1<\/sub> – r2<\/sub>
\nOne circle is inside the other.
\n\u2234 No common tangent.<\/p>\n

Question 3.
\nFind the internal centre of similitude for the circles x\u00b2 + y\u00b2 + 6x – 2y + 1 =0 and x\u00b2 + y\u00b2 – 2x – 6y + 9 = 0.
\nSolution:
\n\"Inter
\nThe internal centre of similitude S divides
\nC1<\/sub> C2<\/sub> internally in the ratio r1<\/sub> : r2<\/sub> = 3 : 1
\nCo-ordinates of S\u00b9 are
\n\"Inter<\/p>\n

Question 4.
\nFind the external centre of similitude for the circles x\u00b2 + y\u00b2 – 2x – 6y + 9 = 0 and x\u00b2 + y\u00b2 = 4.
\nSolution:
\nCentre of the circle C1<\/sub> (1, 3) and C2 (0,0)
\nr1<\/sub> = \\(\\sqrt{1+9-9}\\), r2<\/sub> = 2
\nExternal centre of similitude S\u00b9 divides C1<\/sub> C2<\/sub> externally in the ratio
\nr1<\/sub> : r2<\/sub> = 1 : 2
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\ni) Show that the circles x\u00b2 + y\u00b2 – 6x – 2y + 1 = 0 ; x\u00b2 + y\u00b2 + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
\nSolution:
\nEquations of the circles are
\nS1<\/sub> \u2261 x\u00b2 + y\u00b2 – 6x – 2y + 1 =0
\nS2<\/sub> \u2261 x\u00b2 + y\u00b2 + 2x – 8y + 13 = 0
\n\"Inter
\n\u2234 The circles touch each other externally.
\nThe point of contact P divides AB – internally in the ratio r1<\/sub> : r2<\/sub> = 3 : 2
\nCo-ordinates of P are
\n(\\(\\frac{3(-1)+2.3}{5}\\), \\(\\frac{3.4+2.1}{5}\\) i.e., P(\\(\\frac{3}{5}\\), \\(\\frac{14}{5}\\))
\nEquation of the common tangent is S1<\/sub> – S2<\/sub> = 0
\n-8x + 6y – 12 = 0 (or) 4x – 3y + 6 = 0<\/p>\n

ii) Show that x\u00b2 + y\u00b2 – 6x – 9y + 13 = 0, x\u00b2 + y\u00b2 – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
\nSolution:
\nEquations of the circles are
\nS1<\/sub> \u2261 x\u00b2 + y\u00b2 – 6x – 9y + 13 = 0
\nS2<\/sub> \u2261 x\u00b2 + y\u00b2 – 2x – 16y = 0
\ncentres are A (3, \\(\\frac{9}{2}\\)), B(1, 8)
\n\"Inter
\n\u2234 The circles touch each other internally. The point of contact ‘P’ divides AB externally in the ratio r1<\/sub> : r2<\/sub> = \\(\\frac{\\sqrt{65}}{2}\\) : \u221a65
\n= 1 : 2 Co-ordinates of P are
\n\"Inter
\nP = (5, 1)
\n\u2234 Equation of the common tangent is
\nS1<\/sub> – S2<\/sub> = 0
\n– 4x + 7y +13 = 0
\n4x – 7y – 13 = 0<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the equation of the circle which touches the circle x\u00b2 + y\u00b2 – 2x – 4y – 20 = 0 externally at (5, 5) with radius 5.
\nSolution:
\nx\u00b2 + y\u00b2 – 2x – 4y – 20 = 0 ,
\nC = (1, 2)
\n\"Inter
\nEquation of circle be
\n(x – 9)\u00b2 + (y – 8)\u00b2
\nx\u00b2 + y\u00b2 – 18x – 16y + 120 = 0
\nIf (h, k) is the centre of the required circle (5, 5) is the mid-point of (1, 2) and (h, k).<\/p>\n

Question 3.
\nFind the direct common tangents of the circles.
\nx\u00b2 + y\u00b2 + 22x – 4y – 100 = 0 and x\u00b2 + y\u00b2 – 22x + 4y + 100 = 0.
\nSolution:
\nC1<\/sub> = (-11, 2) C2<\/sub> = (11, -2)
\nr1<\/sub> = \\(\\sqrt{121+4+100}\\) = 15
\nr2<\/sub> = \\(\\sqrt{121+4-100}\\) = 5
\nLet y = mx + c be tangent
\nmx – y + c = 0
\n\u22a5 from (-11, 2) to tangent = 15
\n\u22a5 from (11 ,-2) to tangent = 5
\n\"Inter
\n\"Inter
\nc = -22 m – 4
\n\u2234 y = mx – 22m – 4
\nThis line is a tangent to the second circle
\n\u2234 \\(\\frac{|11m+2-22m-4|}{\\sqrt{m^{2}+1}}\\) = 5
\nSquaring and cross multiplying
\n25(1 + m\u00b2) = (11m + 2 – 22m – 4)\u00b2
\n96m\u00b2 + 44m – 21 = 0
\n\u21d2 96m\u00b2 + 72m – 28m – 21 = 0
\n\"Inter
\n\u21d2 24y = 7x – 250
\n\u21d2 7x – 24y – 250 = 0<\/p>\n

Question 4.
\nFind the transverse common tangents of the circles x\u00b2 + y\u00b2 – 4x – 10y + 28 = 0 and x\u00b2 + y\u00b2 + 4x – 6y + 4 = 0.
\nSolution:
\nC1<\/sub> = (2, 5), C2<\/sub> = (-2, 3)
\n\"Inter
\n= [\\(\\frac{4}{4}\\), \\(\\frac{9}{2}\\)] = [1, \\(\\frac{9}{2}\\)]
\nEquation of the pair transverse of the common tangents is
\n\"Inter<\/p>\n

Question 5.
\nFind the pair of tangents from (4, 10) to the circle x\u00b2 + y\u00b2 = 25.
\nSolution:
\nSS11<\/sub> = S\u00b21<\/sub>
\n(x\u00b2 + y\u00b2 – 25) (16 + 100 – 25)
\n= (4x + 10 y -25)\u00b2
\n91x\u00b2 + 91y\u00b2 – 2275
\n= [16x\u00b2 + 100y\u00b2 + 625 + 80xy – 200x – 500y]
\n75x\u00b2 – 9y\u00b2 – 80xy + 500y + 200x – 2900 = 0<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFind the pair of tangents drawn from (0,0) to x2 + y2 + 10x + 10y + 40 = 0.
\nSolution:
\nS.S11<\/sub> = S\u00b21<\/sub>
\n(x\u00b2 + y\u00b2 + 10x + 10y + 40) (40) = [5x + 5y + 40]\u00b2
\n8(x\u00b2 + y\u00b2 + 10x + 10y + 40) = (x + y + 8)\u00b25
\n8x\u00b2 + 8y\u00b2 + 80x + 80y + 320
\n= 5x\u00b2 + 5y\u00b2 + 10xy + 80x + 80y + 320
\n3x\u00b2 + 3y\u00b2 – 10xy = 0<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nFind the equation of the circle which touches x\u00b2 + y\u00b2 – 4x + 6y – 12 = 0 at (-1, 1) internally with a radius of 2.
\nSolution:
\nC1<\/sub> = (2, -3)
\nr1<\/sub> = \\(\\sqrt{4+9+12}\\) = 5
\n\"Inter
\nLet C2<\/sub> = (h, k)
\npoint of contact – (x, y) = (-1, 1)
\nsince the two circles touch internally
\n\"Inter
\nEquation of a circle with centre (\\(\\frac{1}{5}\\), \\(\\frac{-3}{5}\\)) and radius 2 is given by
\n(x – \\(\\frac{1}{5}\\))\u00b2 + (y + \\(\\frac{1}{5}\\))\u00b2 = 4
\n5x\u00b2 + 5y\u00b2 – 2x + 6y – 18 = 0<\/p>\n

Question 2.
\nFind all common tangents of the following pairs of circles.
\n(i) x\u00b2 + y\u00b2 = 9 and x\u00b2 + y\u00b2 – 16x + 2y + 49 = 0
\nSolution:
\nThe circles are x\u00b2 + y\u00b2 = 9
\nx\u00b2 + y\u00b2 – 16x + 2y + 49 = 0
\nCentre are A (0, 0), B(8, -1)
\n\"Inter
\nThe circles lie outside each other.
\nA (0, 0), B (8, -1)
\nExternal center of similitude S divides AB externally in this ratio 3 : 4
\nCo-ordinates of are (-24, + 3)
\nSuppose m is the slope of the direct common tangents
\ny – 3 = m(x + 24)
\n= mx + 24m
\nmx – y 4- (24m + 3) = 0 ………… (1)
\nThis is a tangent to the circle x\u00b2 + y\u00b2 = 9
\n3 = \\(\\frac{|24m+3|}{\\sqrt{m^{2}+1}}\\)
\n9(m\u00b2 +1 = 9(m\u00b2 + 1) = 9(8m + 1)\u00b2
\n= 64m\u00b2 + 16m + 1
\n63m\u00b2 + 16m = 0
\nm(63m + 16) = 0
\nm = 0 or \\(\\frac{-16}{63}\\).<\/p>\n

Case (i): m = 0
\nSubstituting in (1), equation of the tangent is
\n-y + 3 = 0
\ny – 3 = 0<\/p>\n

Case (ii) : m = \\(\\frac{-16}{63}\\).
\nEquation of the tangent is
\n\"Inter
\nInternal center of similitude S’ divides AB internally in the ratio 3 : 4
\nCo-ordinates of S’ are (\\(\\frac{24}{7}\\), \\(\\frac{-3}{7}\\))
\nEquation of the transverse common tangent is
\n\"Inter
\n7y + 3 = 7mx – 24m
\n7mx – 7y – (24m + 3) = 0 ………… (2)
\nThis is a tangent to the circle x\u00b2 + y\u00b2 = 9
\n3 = \\(\\frac{|24m+3|}{\\sqrt{49m^{2}+49}}=\\frac{3}{\\sqrt{7}} \\frac{|28m+1|}{\\sqrt{m^{2}+1}}\\)
\n49 (m\u00b2 + 1) = (8m + 1)\u00b2
\n49m\u00b2 + 49 = 64m\u00b2 +16m + 1
\n15m\u00b2 + 16m – 48 = 0
\n(3m – 4) (5m + 12) = 0
\nm = \\(\\frac{4}{3}\\) or \\(\\frac{-12}{5}\\)<\/p>\n

Case (i): Substituting in (2), equation of the tangent is
\n\"Inter<\/p>\n

Case (ii): m = \\(\\frac{-12}{5}\\)
\nEquation of the transverse common tangent is ie
\n\"Inter
\n\u2234 Equation of direct common tangents are
\ny – 3 = 0; 16x + 63y + 195 = 0
\nEquation of transverse common tangents are
\n4x – 3y – 15 = 0 and 12x + 5y – 39 = 0<\/p>\n

ii) x\u00b2 + y\u00b2 + 4x + 2y – 4 = 0 and x\u00b2 + y\u00b2 – 4x – 2y + 4 = 0
\nSolution:
\nr1<\/sub> = \\(\\sqrt{4+1+4}\\) = 3, r2<\/sub> = \\(\\sqrt{4+1+4}\\) = 1
\nExternal center of similitude S divides AB externally in the ratio 3 : 1
\n3 : 1 A (-2,-1), S’ B (2, 1) S
\nCo-ordinates of S are
\n(\\(\\frac{3.2-(-2)1}{3-1}\\), \\(\\frac{3.1-1(-1)}{3-1}\\)) = (4, 2)
\nSuppose m is the slope of the tangent Equation of the tangent can be taken as
\ny – 2 = m(x – 4)
\n= mx – 4m
\nmx – y + (2 – 4m) = 0 ………… (1)
\nThis is a tangent to the circle
\nx\u00b2 + y\u00b2 + 4x + 2y – 4 = 0
\n\"Inter
\nSquaring and cross-multiplication is
\n(1 – 2m)\u00b2 = (m\u00b2 + 1)
\n4m\u00b2 – 4m + 1 = m\u00b2 + 1
\n3m\u00b2 – 4m = 0
\nm(3m-4) = 0
\nm,= 0 or = \\(\\frac{4}{3}\\)
\nm = 0
\nSubstituting in (1), equation of the tangent is y + 2 = 0 or y -2 = 0
\nm = \\(\\frac{4}{3}\\)
\nSubstituting in (1), equation of the tangent is
\n\\(\\frac{4}{3}\\)x – y + (2 – \\(\\frac{16}{3}\\)) = 0
\n\u21d2 \\(\\frac{4}{3}\\)x – y –\\(\\frac{10}{3}\\) = 0
\n4x – 3y – 10 = 0
\nInternal centre of similitude S’ divides AB internally in the ratio 3 : 1
\nCo-ordinates of S’ are
\n(\\(\\frac{6-2}{3+1}\\), \\(\\frac{3-1}{3+1}\\)x) = (1, \\(\\frac{1}{2}\\))
\nEquation of the tangent can be taken as
\ny – \\(\\frac{1}{2}\\) = m(x – 1)
\n= mx – m
\nmx – y + (\\(\\frac{1}{2}\\) – m) = 0 ………. (2)
\nThis is a tangent to the circle
\nx\u00b2 + y\u00b2 + 4x + 2y – 4 = 0
\n\"Inter
\nSquaring and cross-multiply is
\n(1 – 2m)\u00b2 = 4(m\u00b2 + 1)
\n1 + 4m\u00b2 – 4m = 4m\u00b2 + 4
\nHere then value of m is a, so that the tangent is a vertical line, equation of the tangent is
\nx = \\(\\frac{1}{2}\\)
\nx – 1 = 0
\n4m + 3 = 0
\nm = \\(\\frac{-3}{4}\\)
\nSubstituting in (2), equation of the tangent is
\n\\(\\frac{-3}{4}\\)x – y + (\\(\\frac{1}{2}+\\frac{3}{4}\\)) = 0
\n-3x – 4y + 5 = 0
\n3x + 4y – 5 = 0
\nEquation of direct common tangents are
\ny – 2 = 0, 4x – 3y – 10 = 0
\nEquation of transvere common tangents are
\nx – 1 = 0, 3x + 4y – 5 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the pair of tangents drawn from (3, 2) to the circle x\u00b2 + y\u00b2 – 6x + 4y – 2 = 0.
\nSolution:
\nS.S11<\/sub> = S\u00b21<\/sub>
\n(x\u00b2 + y\u00b2 – 6x + 4y – 2)
\n(9+ 4 – 6 \u00d7 3 + 4 \u00d7 2 – 2)
\n= (3x + 2y – 3(x + 3) + 2(y + 2) – 2)\u00b2
\n(x\u00b2 + y\u00b2 – 6x + 4y – 2) = (4y – 7)\u00b2
\nx\u00b2 + y\u00b2 – 6x + 4y – 2 = 16y\u00b2 – 56y + 49
\nx\u00b2 – 15y\u00b2 – 6x + 60y – 51 = 0<\/p>\n

Question 4.
\nFind the pair of tangents drawn from (1, 3) to the circle x\u00b2 + y\u00b2 – 2x + 4y – 11 = 0 and also find the angle between them.
\nSolution:
\nS.S11<\/sub> = S\u00b21<\/sub>
\n(x\u00b2 + y\u00b2 – 2x + 4y -11) (1 + 9-2 + 12-11) = [x + 3y – 1 (x + 1) + 2 (y + 3)- 11]\u00b2
\n(x\u00b2 + y\u00b2 – 2x + 4y – 11)9 = [5y-6]\u00b2
\n9x\u00b2 + 9y\u00b2 – 18x + 36y – 99
\n25y\u00b2+ 36 – 60 y
\n9x\u00b2 – 16y\u00b2 – 18x + 96y – 135 = 0
\n\"Inter<\/p>\n

Question 5.
\nFind the pair of tangents from the origin to the circle x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0 and hence deduce a condition for these tangents to be perpendicular.
\nSolution:
\nS.S11<\/sub> = S\u00b21<\/sub>
\n(x\u00b2 + y\u00b2 + 2gx + 2fy + c) (c) = [gx + fy + c]\u00b2
\n= g\u00b2x\u00b2 +f\u00b2y\u00b2+ 2gfxy + 2gcx + 2fyc + c\u00b2
\n(gx + fy)\u00b2 = c(x\u00b2 + y\u00b2)
\ng\u00b2x\u00b2 + f\u00b2y\u00b2 + 2fg xy = cx\u00b2 + cy\u00b2
\n(g\u00b2 – c)x + 2fgxy + (f\u00b2 – c)y\u00b2 = 0
\nco-efficient of x\u00b2 + co-efficient of y\u00b2 = 0
\ng\u00b2 – c + f\u00b2 – c = 0
\ng\u00b2 + f\u00b2 = 2c<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFrom a point on the circle x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0, two tangents are drawn to the circle x\u00b2 + y\u00b2 + 2gx + 2fy + c sin\u00b2 \u03b1 + (g\u00b2 + f\u00b2) cos\u00b2 \u03b1 = 0 (0 < a < \u03c0\/2). Prove that the angle between them is 2\u03b1.
\nSolution:
\n[x\u00b21<\/sub> + y\u00b21<\/sub> + 2gx1<\/sub> + 2fy1<\/sub> + c sin\u00b2 \u03b1 + (g\u00b2 + f\u00b2) cos\u00b2 \u03b1] (S)
\n= (xx1<\/sub> + yy1<\/sub> + g(x + x1<\/sub>) + f(y + y1<\/sub>) + c sin\u00b2 \u03b1 + (g\u00b2 + f\u00b2) cos\u00b2 \u03b1)\u00b2
\n[(-c + c sin\u00b2 \u03b1)+ (g\u00b2 + f\u00b2) cos\u00b2 \u03b1]S
\n= (x (x1<\/sub> + g) + y (y1<\/sub> + f) + gx1<\/sub> + fy1<\/sub> + c sin\u00b2 \u03b1 + (g\u00b2 + f\u00b2) cos\u00b2 \u03b1)\u00b2
\n[cos\u00b2 \u03b1 (g\u00b2 + f\u00b2 – c)] S
\n= [x (x1<\/sub> + g) + y (y1<\/sub> + f) + gx1<\/sub> + fy1<\/sub> + c sin\u00b2 \u03b1 + (g\u00b2 + f\u00b2)cos\u00b2 \u03b1
\nLet g\u00b2 + f\u00b2 – c = r\u00b2
\n= [x (x1<\/sub> + g) + y (y1<\/sub> + f) + gx1<\/sub> + fy1<\/sub> + gx1<\/sub> + fy1<\/sub> + c + (cos\u00b2 \u03b1).r\u00b2)\u00b2
\nCoefficient of x\u00b2 is r\u00b2 cos\u00b2 \u03b1 – (x1<\/sub> + g)\u00b2
\nCoefficient of y\u00b2 is r\u00b2 cos2 \u03b1 – (y1<\/sub> + f)\u00b2
\nCoefficient of xy is
\nh. cos\u00b2 \u03b1 r\u00b2 – 2 (x1<\/sub> + g) (y1<\/sub> + f)
\n\"Inter
\ncos \u03b8 = cos 2 \u03b1
\n\u03b8 = 2\u03b1
\nHence proved.<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(e) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(e) I. Question 1. Discuss the relative position of the following pair of circles. i) x\u00b2 + y\u00b2 – 4x – …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35510"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35510"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35510\/revisions"}],"predecessor-version":[{"id":35511,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35510\/revisions\/35511"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35510"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35510"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35510"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}