Suggested languages for you:

Americas

Europe

Problem 15

The equation of tangents drawn from the origin to the circle $x^{2}+y^{2}-2 r x-2 h y+h^{2}=0$, are : (a) \(x=0\) (b) \(y=0\) (c) \(\left(h^{2}-r^{2}\right) x-2 r h y=0\) (d) \(\left(h^{2}-r^{2}\right) x+2 r h y=0\)

Expert verified

The correct equation of the tangent to the given circle from the origin is \((h^2 - r^2)x + 2rhy = 0\). Therefore, option (c) is the correct choice.

What do you think about this solution?

We value your feedback to improve our textbook solutions.

- Access over 3 million high quality textbook solutions
- Access our popular flashcard, quiz, mock-exam and notes features
- Access our smart AI features to upgrade your learning

Chapter 1

If the sum of the distances of a point from two perpendicular lines in the plane is $\begin{array}{ll}\text { 1, then its locus is : } & \text { (b) a circle }\end{array}$ (a) a straight line (c) a parallelogram (d) an ellipse

Chapter 1

Find the equation of that diameter of the circle \(x^{2}+y^{2}-6 x+2 y-8=0\) which passes through the origin.

Chapter 1

If \(A(a, a), B(-a,-a)\) are two vertices of an equilateral triangle, then its third vertex is: (a) \(\left(\frac{a \sqrt{3}}{2},-\frac{a \sqrt{3}}{2}\right)\) (b) \((-a \sqrt{3}, a \sqrt{3})\) (c) \((a \sqrt{3},-a \sqrt{3})\) (d) \((-a \sqrt{3},-a \sqrt{3})\)

Chapter 1

The circle \(x^{2}+y^{2}=4\) cuts the circle \(x^{2}+y^{2}+2 x+3 y-5=0\) in \(A\) and \(B\), the centre of the circle \(A B\) as diameter is : (a) \((0,0)\) (b) \(\left(\frac{2}{13}, \frac{3}{13}\right)\) (c) \(\left(\frac{4}{13}, \frac{6}{13}\right)\) (d) \((2,-1)\)

Chapter 1

If \(A(\cos \theta, \sin \theta), B(\sin \theta, \cos \theta), C(1,2)\) are the vertices of a \(\triangle A B C\). Find the locus of its centroid if \(\theta\) varies.

The first learning app that truly has everything you need to ace your exams in one place.

- Flashcards & Quizzes
- AI Study Assistant
- Smart Note-Taking
- Mock-Exams
- Study Planner