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 […]
Solution : Let us assume that the coordinates of the center of the circle are C(h,k) and its radius is r. Now, since the circle touches X-axis at (1,0), hence its radius should be equal to ordinate of center. \(\implies\) r = k Hence, the equation of circle is \((x – h)^2 + (y –
Solution : Given the equation of ellipse is \(x^2\over 16\) + \(y^2\over 9\) = 1 Here, a = 4, b = 3, e = \(\sqrt{1-{9\over 16}}\) = \(\sqrt{7\over 4}\) \(\therefore\) foci is (\(\pm ae\), 0) = (\(\pm\sqrt{7}\), 0) \(\therefore\) Radius of the circle, r = \(\sqrt{(ae)^2+b^2}\) r = \(\sqrt{7+9}\) = \(\sqrt{16}\) = 4 Now, equation
Solution : Let the equation of circle be \((x-3)^2 + (y-0)^2 + \lambda y\) = 0 As it passes through (1, -2) \(\therefore\) \((1-3)^2 + (-2)^2 + \lambda (-2)\) = 0 \(\implies\) 4 + 4 – 2\(\lambda\) = 0 \(\implies\) \(\lambda\) = 4 \(\therefore\) Equation of circle is \((x-3)^2 + y^2 + 4y\) = 0
Solution : Let coordinates of the center of T be (0, k). Distance between their center is k + 1 = \(\sqrt{1 + (k – 1)^2}\) where k is radius of circle T and 1 is radius of circle C, so sum of these is distance between their centers. \(\implies\) k + 1 = \(\sqrt{k^2 +
Solution : Family of circles is \(x^2 + y^2 – 2x – 4y + 1\) + \(\lambda\)(\(x^2 + y^2 – 1\)) = 0 (1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) – 2x – 4y + (1 – \(\lambda\))) = 0 \(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y +
Solution : Pair of normals are (x + 2y)(x + 3) = 0 \(\therefore\) Normals are x + 2y = 0, x + 3 = 0 Point of intersection of normals is the center of the required circle i.e. \(C_1\)(-3,3/2) and center of the given circle is \(C_2\)(2,3/2) and radius \(r^2\) = \(\sqrt{4 + {9\over
Solution : Since the normal to the circle always passes through the center, so the equation of the normal will be the line passing through (5,6) & (\(5\over 2\), -1) i.e. y + 1 = \(7\over {5/2}\)(x – \(5\over 2\)) \(\implies\) 5y + 5 = 14x – 35 \(\implies\) 14x – 5y – 40 =
Solution : Area = 49π π\(r^2\) = 49π r = 7 Now find the coordinates of center of circle by solving the given two equations of diameter. By solving the above equation through elimination method we get, x = 1 and y =-1 which are the coordinates of center of circle. Now the general equation
