The greatest distance of the point P (10,7) from the circle x^ 2 +y^ 2 -4 x-2 y-20=0 is
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
The greatest distance of the point \(\mathrm{P}(10,7)\) from the circle \(x^{2}+y^{2}-4 x-2 y-20=0\) is
- 10 units
- 15 units
- 5 units
- None of these
Answer
(B) 15 units
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The equation of the tangents to the circle x 2 + y 2 = 4, which are parallel to x + 2y + 3 = 0, are
- The equation of the common tangent touching the circle (x-3)^ 2 +y^ 2 =9 and the parabola ^ 2 = 4 above the…
- The line 3x – 2y = k meets the circle x 2 + y 2 = 4r 2 at only one point, if k 2 is
- Any point on the parabola whose focus is (0, 1) and the directrix is x + 2 = 0 is given by
- If the straight line through the point P(3, 4) makes an angle 6 with x-axis and meets the line 12x + 5y + 10…
- The foci of a hyperbola coincide with the foci of the ellipse x 2 /25 + y 2 /9 = 1 . If eccentricity of the…
- The equation of the tangent to the parabola ^ 2 = 4 a at the point (3, 2) is,
- The point P is equidistant from A(1, 3), B(- 3, 5) and C(5, - 1), then PA is equal to
More Co-ordinate Geometry questions · Browse all practice questions