Circle Questions

What is the equation of pair of tangents to a circle ?

Solution : Let the equation of circle S = \(x^2\) + \(y^2\) = \(a^2\) and P(\(x_1,y_1\)) is any point outside the circle. From the point we can draw two real and distinct tangent and combine equation of pair of tangents is – (\(x^2\) + \(y^2\) – \(a^2\))(\({x_1}^2\) + \({y_1}^2\) – \(a^2\)) = \(({xx_1 + yy_1 …

What is the equation of pair of tangents to a circle ? Read More »

What is the length of tangent to a circle formula from an external point ?

Solution : The length of tangent drawn from point (\(x_1,y_1\)) outside the circle S = \(x^2 + y^2 + 2gx + 2fy + c\) = 0 is, \(\sqrt{S_1}\) = (\(\sqrt{{x_1}^2 + {y_1}^2 + 2gx_1 + 2fy_1 + c}\)) Similar Questions What is the Director Circle of a Circle ? What are the Intercepts cut by …

What is the length of tangent to a circle formula from an external point ? Read More »

What is the parametric equation of circle ?

Solution : 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. Similar Questions What is the Director Circle of a Circle ? Find the number of common tangents to the circles \(x^2 + y^2\) = 1 …

What is the parametric equation of circle ? Read More »

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 …

Find the number of common tangents to the circles \(x^2 + y^2\) = 1 and \(x^2 + y^2 – 2x – 6y + 6\) = 0. Read More »

What are Orthogonal Circles and Condition of Orthogonal Circles.

Solution : 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 is cos\(\theta\) = |\(2{g_1}{g_2} + 2{f_1}{f_2} – {c_1} – {c_2}\over {2\sqrt{{g_1}^2 + {f_1}^2 -c_1}}{\sqrt{{g_1}^2 + {f_1}^2 …

What are Orthogonal Circles and Condition of Orthogonal Circles. Read More »

What is the formula for the angle of intersection of two circles ?

Solution : 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 \(S_1\) = \({x}^2 + {y}^2 + 2{g_1}x + 2{f_1}y + {c_1}\) = 0 and \(S_2\) = \({x}^2 + {y}^2 + 2{g_2}x + 2{f_2}y + …

What is the formula for the angle of intersection of two circles ? Read More »

The length of the diameter of the circle which touches the X-axis at the point (1,0) and passes through the point (2,3) is

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 length of the diameter of the circle which touches the X-axis at the point (1,0) and passes through the point (2,3) is Read More »

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 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 Read More »