Circle Questions

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