If the points (2,0) , (3,2) , (5,4) , and (t,0) are con-cyclic , then find the value of t.

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Please find below the solution to the asked query:

As given points are con-cyclicx2+y2+2gx+2fy+c=0 ;iAs given points are Put 2,0 in i4+0+4g+0+c=0c=-4g-4 ;iiPut 3,2 in i9+4+6g+4f+c=013+6g+4f+c=013+6g+4f-4g-4 =0  Using ii2g+4f+9=02g=-4f-9 ;iiiPut 5,4 in i25+16+10g+8f+c=041+10g+8f-4g-4=0 Using ii6g+8f+37=03-4f-9+8f+37=0 Using iii-12f-27+8f+37=0-4f+10=04f=102f=5Put this value in iii2g=-25-92g=-19Put this value in ic=-2-19-4c=34Putting obtained values in i, we get:x2+y2-19x+5y+34=0Put t,0 in above equation:t2+0-19x+0+34=0t2-19x+34=0t2-2x-17x+34=0tt-2-17t-2=0t-2t-17=0Hence t=2 or t=17

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