Circle Questions

Find the number of common tangents to the circles $$x^2 + y^2$$ = 1 and $$x^2 + y^2 – 2x – 6y + 6$$ = 0.

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 – … The equation of the circle passing through the foci of the ellipse \(x^2\over 16$$ + $$y^2\over 9$$ = 1 and having center at (0, 3) is

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 …

The circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point

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 : 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 … Find the equation of the normal to the circle \(x^2 + y^2 – 5x + 2y -48$$ = 0 at the point (5,6).

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 = …

If the lines 3x-4y-7=0 and 2x-3y-5=0 are two diameters of a circle of area 49π square units, then what is the equation of the circle?

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 …