slopes of tangents through (7,1) to the circle x^2+y^2=25 satisfy the equation
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x2+y2=25⇒(x−0)2+(y−0)2=52Hence centre is (0,0) and radius is a=5 unitsAny tangent of m slope to above cirlce isy=mx±a1+m2−−−−−−√a=5⇒y=mx±51+m2−−−−−−√Tangent passes through (7,1)⇒1=7m±51+m2−−−−−−√⇒1−7m=±51+m2−−−−−−√Square both sides⇒(1−7m)2=25(1+m2)⇒1+49m2−14m=25+25m2⇒24m2−14m−24=0⇒12m2−7m−12=0
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x2+y2=25⇒(x−0)2+(y−0)2=52Hence centre is (0,0) and radius is a=5 unitsAny tangent of m slope to above cirlce isy=mx±a1+m2−−−−−−√a=5⇒y=mx±51+m2−−−−−−√Tangent passes through (7,1)⇒1=7m±51+m2−−−−−−√⇒1−7m=±51+m2−−−−−−√Square both sides⇒(1−7m)2=25(1+m2)⇒1+49m2−14m=25+25m2⇒24m2−14m−24=0⇒12m2−7m−12=0
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