Question

Let A(-7,0) and B(7,0) be two fixed points on a circle. Find the locus of a moving
point P at which AB subtend a right angle.​

Answer 1

Step-by-step explanation:

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2π

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2πThen, slope of AP× slope of BP=−1

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2πThen, slope of AP× slope of BP=−1⇒h−x1k−y1×h−x2k−y2=−1

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2πThen, slope of AP× slope of BP=−1⇒h−x1k−y1×h−x2k−y2=−1⇒(h−x1)(h−x2)+(k−y1)(k−y2)=0

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2πThen, slope of AP× slope of BP=−1⇒h−x1k−y1×h−x2k−y2=−1⇒(h−x1)(h−x2)+(k−y1)(k−y2)=0Hence, locus of (h,k) is

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2πThen, slope of AP× slope of BP=−1⇒h−x1k−y1×h−x2k−y2=−1⇒(h−x1)(h−x2)+(k−y1)(k−y2)=0Hence, locus of (h,k) is(x−x1)(x−x2)+(y−y1)(y−y2)=0

Let A(x1,y1) and B(x2,y2) be two fixed points and P(h,k) be a variable point such that∠APB=2πThen, slope of AP× slope of BP=−1⇒h−x1k−y1×h−x2k−y2=−1⇒(h−x1)(h−x2)+(k−y1)(k−y2)=0Hence, locus of (h,k) is(x−x1)(x−x2)+(y−y1)(y−y2)=0which is a circle having AB as diameter.