Question:
Find the centre and radius of the circle $x^{2}+y^{2}-4 x-8 y-45=0$
Solution:
The equation of the given circle is $x^{2}+y^{2}-4 x-8 y-45=0$.
$x^{2}+y^{2}-4 x-8 y-45=0$
$\Rightarrow\left(x^{2}-4 x\right)+\left(y^{2}-8 y\right)=45$
$\Rightarrow\left\{x^{2}-2(x)(2)+2^{2}\right\}+\left\{y^{2}-2(y)(4)+4^{2}\right\}-4-16=45$
$\Rightarrow(x-2)^{2}+(y-4)^{2}=65$
$\Rightarrow(x-2)^{2}+(y-4)^{2}=(\sqrt{65})^{2}$, which is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $h=2, k=4$, and $r=\sqrt{65}$.
Thus, the centre of the given circle is $(2,4)$, while its radius is $\sqrt{65}$.
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- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Wave
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- Atoms
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