BITSAT Mathematics Question (2022) — Solution

Step-by-step solution

Let P (t, 4 t-20 5 ) be a point on the line 4 x-5 y=20. Then the chord of contact of tangents drawn from P to the circle x^2+y^2=9 is t x+ ( 4 t-20 5 ) y=9 (i) Let (h, k) be the mid-point of this chord of contact, then its equation is also h x+k y=h^2+k^2 (ii) [ . using .T=S^ ] Clearly Eqs. (i) and (ii) represent the same line aligned & t h = 4 t-20 5 k = 9 h^2+k^2 \\ & t h = 9 h^2+k^2 and t h = 4 t-20 5 k \\ & t= 9 h h^2+k^2 and t= 20 h 4 h-5 k \\ & 9 h h^2+k^2 = 20 h 4 h-5 k \\ & h \ 20 (h^2+k^2 )-36 h+45 k \ =0 \\ & x=0, or [20 (x^2+y^2 )-36 x+45 y ]=0 aligned