Parametric Equation of Circle (i) The parametric equation of circle \(x^2 + y^2\) = \(r^2\) are x = rcos\(\theta\), y = rsin\(\theta\) ; \(\theta\) \(\in\) [0,2\(\pi\)) and (rcos\(\theta\), rsin\(\theta\)) are called parametric coordinates. Also Read : General Equation of the Circle – Formula and Examples (ii) The parametric equation of circle \((x – h)^2 + …
Here you will learn what is the equation of director circle of circle with proof. Let’s begin – Director Circle of a Circle The equation of director circle is \(x^2\) + \(y^2\) = 2\(a^2\). Proof : The locus of the point of intersection of two perpendicular tangents to the circle is called director circle. Let …
Director Circle of a Circle – Equation and Proof Read More »
Solution : Let \(C_1\) be the center of circle \(x^2 + y^2\) = 1 i.e. \(C_1\) = (0, 0) And \(C_2\) be the center of circle \(x^2 + y^2 – 2x – 6y + 6\) = 0 i.e. \(C_2\) = (1, 3) Let \(r_1\) be the radius of first circle and \(r_2\) be the radius …
Here you will learn what are orthogonal circles and condition of orthogonal circles. Let’s begin – Orthogonal Circles Let two circles are \(S_1\) = \({x}^2 + {y}^2 + 2{g_1}x + 2{f_1}y + {c_1}\) = 0 and \(S_2\) = \({x}^2 + {y}^2 + 2{g_2}x + 2{f_2}y + {c_2}\) = 0.Then Angle of intersection of two circles …
Orthogonal Circles and Condition of Orthogonal Circles Read More »
Here you will learn what is the formula for the angle of intersection of two circles. Let’s begin – Angle of Intersection of Two Circles Formula The angle between the tangents of two circles at the point of intersection of the two circles is called angle of intersection of two circles. If two circles are …
Solution : The given expression is \((3x – {x^3\over 6})^7\). Here n = 7, which is an odd number. By using middle terms in binomial expansion formula, So, \(({7 + 1\over 2}\)) th and \(({7 + 1\over 2} + 1)\) th i.e. 4th and 5th terms are two middle terms. Now, \(T_{4}\) = \(T_{3 + …
Find the middle term in the expansion of \((3x – {x^3\over 6})^7\). Read More »
Solution : Here, n = 20, which is an even number. So, \({20\over 2} + 1\)th term i.e. 11th term is the middle term. Hence, the middle term = \(T_{11}\) \(T_{11}\) = \(T_{10 + 1}\) = \(^{20}C_{10}\) \(({2\over 3}x^2)^{20 – 10}\) \((-{3\over 2x})^{10}\) = \(^{20}C_{10} x^{10}\) Similar Questions Find the middle term in the expansion …
Find the middle term in the expansion of \(({2\over 3}x^2 – {3\over 2x})^{20}\). Read More »
Solution : We know that the (r + 1)th term or general term in the expansion of \((x + a)^n\) is given by \(T_{r + 1}\) = \(^{n}C_r x^{n – r} a^r\) In the expansion of \(({x\over a} – {3a\over x^2})^{12}\), we have \(T_{9}\) = \(T_{8 + 1}\) = \(^{12}C_8 ({x\over a})^{12 – 8} ({-3a\over …
Find the 9th term in the expansion of \(({x\over a} – {3a\over x^2})^{12}\). Read More »
Solution : We know that the (r + 1)th term or general term in the expansion of \((x + a)^n\) is given by \(T_{r + 1}\) = \(^{n}C_r x^{n – r} a^r\) In the expansion of \((2x^2 + {1\over x})^{12}\), we have \(T_{10}\) = \(T_{9 + 1}\) = \(^{12}C_9 (2x^2)^{12 – 9} ({1\over x})^9\) \(\implies\) …
Find the 10th term in the binomial expansion of \((2x^2 + {1\over x})^{12}\). Read More »
Solution : We have, \((1.01)^{1000000}\) – 10000 = \((1 + 0.01)^{1000000}\) – 10000 By using binomial theorem, = \(^{1000000}C_0\) + \(^{1000000}C_1 (0.01)\) + \(^{1000000}C_2 (0.01)^2\) + …… + \(^{1000000}C_{1000000} (0.01)^{1000000}\) – 10000 = (1 + 1000000(0.01) + other positive terms) – 10000 = (1 + 10000 + other positive terms) – 10000 = 1 + …
