Let y=x+2,4y=3x+6 and 3y=4x+1 be three tangent lines to the circle (x−h)2+(y−k)2=r2. Then h+k is equal to:
Explanation
Solution
Center (h,k) ki distance teeno lines se r (radius) ke barabar hogi.
Lines: L1:x−y+2=0, L2:3x−4y+6=0, L3:4x−3y+1=0.
Distance formula: 2∣h−k+2∣=5∣3h−4k+6∣=5∣4h−3k+1∣.
Solving 5∣3h−4k+6∣=5∣4h−3k+1∣ gives h+k=5 or h−k=1.
h−k=1 and 2∣h−k+2∣=5∣3h−4k+6∣ checking leads to h+k=5.
Correct Option: (4)
Explanation
Solution
Center (h,k) ki distance teeno lines se r (radius) ke barabar hogi.
Lines: L1:x−y+2=0, L2:3x−4y+6=0, L3:4x−3y+1=0.
Distance formula: 2∣h−k+2∣=5∣3h−4k+6∣=5∣4h−3k+1∣.
Solving 5∣3h−4k+6∣=5∣4h−3k+1∣ gives h+k=5 or h−k=1.
h−k=1 and 2∣h−k+2∣=5∣3h−4k+6∣ checking leads to h+k=5.
Correct Option: (4)
