Consider the circle x^2+y^2-4 x-2 y+c=0 whose centre is A(2,1) . If the point P(10,7) is such that the line segment P A meets the circle in Q with P Q=5 , then c is equal to

Step-by-step solution

Given equation of circle is x^2+y^2-4 x-2 y+c=0 whose centre is A(2,1) . Now, A P= (2-10)^2+(1-7)^2 aligned & = (-8)^2+(-6)^2 = 64+36 = 100 & =10 aligned A Q=A P-P Q=10-5=5 So, Q is the mid-point of AP = ( 10+2 2 , 7+1 2 )=(6,4) Since, Q lies on a circle. array rlrl & & 6^2+4^2-4(6)-2(4)+c & =0 & & 36+16-24-8+c & =0 & & 20+c & =0 & c & =-20 array