let s1 x 2 y 2 4x 8y 4 0

let s1 x 2 y 2 4x 8y 4 0 Step by step video image solution for let S 1 x 2 y 2 4x 8y 4 0 and S 2 be its image in the line y x find the equation of circle touching y x at 1 1 and its radical axes with S 2 passes through the centre of S 1 by Maths experts to help you in doubts scoring excellent marks in Class 11 exams

Let S1 x2 y2 4x 8y 4 0 and S2 be its image in the line y x find the equation of circle touching y x at 1 1 and its radical axes with S2 passes through the centre S1 See answers Advertisement GauravSaxena01 Solution S x 7 2 2 y 11 2 2 81 2 Please check Attachment file I hope it s help you Advertisement arindam999 Solution Verified by Toppr S 1 x 2 y 2 4x 8y 4 0 S 2 image of S 1 in line y x Centre of S 1 is 2 4 r sqrt 2 2 4 2 4 4 left because x 2 y 2 2gx 2fy c 0 centre h x g f right r g 2 f 2 c S 1 centre 2 4 r 1 4 S 2 is image so centre is 4 2 r 2 4 Eqaution of S 2 x h 2 y k 2 r 2 x 4 2 y 2 2 16

let s1 x 2 y 2 4x 8y 4 0

let-s1-x-2-y-2-4x-8y-4-0-and-s2-be-its-image-in-the-line-y

let s1 x 2 y 2 4x 8y 4 0
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equation-of-tangent-to-circle-x-2-y-2-8x-10x-128-0-ib-math-gcse-youtube

Equation Of Tangent To Circle X 2 y 2 8x 10x 128 0 IB Math GCSE YouTube
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4x-8y-16

4x 8y 16
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314 views 5 years ago To ask Unlimited Maths doubts download Doubtnut from goo gl 9WZjCW let S 1 x 2 y 2 4x 8y 4 0 and S 2 be its image in the The correct answer is Centre of circle S1 2 4 Centre of circle S2 4 2 Radius of circle S1 radius of circle S2 4 equation of cricle S2 x 4 2 y 2 2 16 x2 y2 8x 4y 4 0

U can do this by solving the eqns y x and the eqn woyh slope 1 and passing through center of S1 then asssume radical axis eqn be y mx c and pass it through center of S1 as given One more information is given apply perpendicular distance formulae for eqn given from center of the circle S1 Detailed solution Correct option is C Centre of S 1 2 4 Radius of S 1 r a d i u s o f S 2 4 Centre of circle S 2 4 2 S 2 x 4 2 y 2 2 16 x 2 y 2 8 x 4 y 4 0 1 Equation of the circle touching y x at 1 1 can be taken as

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Popular Problems Precalculus Write in Standard Form x 2 y 2 4x 8y 4 0 x2 y2 4x 8y 4 0 x 2 y 2 4 x 8 y 4 0 Subtract 4 4 from both sides of the equation x2 y2 4x 8y 4 x 2 y 2 4 x 8 y 4 Complete the square for x2 4x x 2 4 x Tap for more steps x 2 2 4 x 2 2 4 VIDEO ANSWER The situation is off the circle as X is scared There s also whole scare off We can change it to X squared There s also whole scare off The standard recreation off the circle is equal to the I minus two being too scared to compete

Let the circles S1 x2 y2 4x 8y 4 0 and S2 be its image in the line y x the equation of the circle touching y x at 1 1 and orthogonal to S2 is a x2 y2 x 5y 2 0 b x2 y2 2 c x2 y2 x 5y 2 0 d x 3 2 y 2 2 5 Correct answer is option A Free system of linear equations calculator solve system of linear equations step by step

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let s1 x 2 y 2 4x 8y 4 0 - Solve by Substitution Calculator Step 1 Enter the system of equations you want to solve for by substitution The solve by substitution calculator allows to find the solution to a system of two or three equations in both a point form and an equation form of the answer Step 2 Click the blue arrow to submit