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Find the equation to the circle which passes through the points (1,-2)(4,-3) and which has its centre on the straight line 3x+4y=7. I am not getting the answer. Answer is 15x²+15y²-94x+18y+55

Find the equation to the circle which passes through the points (1,-2)(4,-3) and which has its centre on the straight line 3x+4y=7.  I am not getting the answer.
Answer is 15x²+15y²-94x+18y+55

Grade:11

1 Answers

Kushagra dutt
24 Points
6 years ago
Let the centre of the circle be (h,k)which Satisfy the equation of the line which passes through the the centre i.e. 3h+4k=7...........(eq.1)Now from the two given point (1,-2) and (4,-3) equate the radius by using distance formula and the final equation you get is3h-k=11...............(eq.2)Now by equating eq.1 and eq.2You will geth=17/5 and k=-4/5Now find the radius by using distance formula i.e. radius=under root 180/25Now putting the values of h,k and radius in the equation of circle (x-h)^2+(y-k)^2=(radius)^2You will get the equation of circle as25x^2+25y^2-170x +40y +125=0

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