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  • Lines <img alt="\mathrm{5 x+12 y-10=0 \text { and } 5 x-12 y-40=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B5%20x&plus;12%20y-10%3D0%20%5Ctext%20%7B%20and%20%7D%20

Lines \mathrm{5 x+12 y-10=0 \text { and } 5 x-12 y-40=0} touch a circle \mathrm{C_{1}} of diameter 6. If the centres of \mathrm{C_{1}} lies in the first quadrant find the equation of the circle \mathrm{C_{2}} which is concentric with \mathrm{C_{1}}and cuts intercepts of length 8 on these lines.

 

Option: 1

\mathrm{(x+5)^2+(y+2)^2=25}


Option: 2

(x+2)^2+(y-5)^2=16


Option: 3

\mathrm{(x+4)^2+(y-3)^2=25}


Option: 4

\mathrm{(x-5)^2+(y-2)^2=25}


Answers (1)

best_answer

Let C(x, y) be the center of the circles. 

Drawing perpendiculars to the tangents (5x – 12y – 40)/13 = ± 3 ⇒ 5x –12y – 40 = ±39……(i) 

Similarly for the second line, we have 5x – 12y – 10 = ± 39 …..(ii) 

There are four possible pairs of equations from (i) and (ii)

∴ To get positive values for x and y we must take 

5x – 12y – 40 = –39 and 5x + 12y – 10 = 39

Solving we get, x = 5 and y = 2 , C ≡ (5, 2) lies in the first quadrant. 

Hence the circle is \mathrm{(x-5)^2+(y-2)^2=r^2}

Now from the figure, \mathrm{r^2=4^2+3^2=25 \text {. Hence } C_2 \text { is }(x-5)^2+(y-2)^2=25}

 

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manish

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