If the mid point of the chord intercepted by the circle x^2+y^2-8 x+10 y+5=0 on the line 2 x+y+2=0 is ( h , k ) , then k +4 ~h =

Step-by-step solution

Given the equation of circle aligned & x^2+y^2-8 x+10 y+5=0 & Centre =(4,-5) aligned Now, equation of chord is 2 x + y +2=0 since ( h , k ) is mid point of chord 2 ~h + k +2=0 ... (i) Now, slope of AB Slope of PC =-1 aligned & -2 ( k +5 ~h -4 )=-1 & 2 k +10= h -4 h -2 k =14 .... (ii) aligned On solving (i) & (ii), we get h =2, k =-6 Now k+4 h=-6+8=2