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Question

If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x−2)2+(y−3)2=25 at the point (5,7) is A, then 24A is equal to


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Solution

Step 1: Draw a schematic diagram diagram with the given date in the question

From the given equation of circle (x−2)2+(y−3)2=25, the centre is at 2,3 and its radius is 5units, then

Step 2: Find the equation of normal,

From point 2,3 and (5,7), the equation of normal is

y−3=7−35−2(x−2)y−3=43(x−2)........(i)

When y=0, then

−9=4x−8−9+8=4xx=−14

Then the point of intersection with the x-axis, that is point Q-14,0

Step 3: Find the equation of tangent,

we know that the product of slopes of normal and tangent is equal to -1, as both are perpendicular to each other.

From equation (i) the slope of normal, m1=43,

we know m1×m2=-1, then m2=-34,

Using the slope m2=-34 and a point (5,7), the equation of tangent can be written as

y−7=-34(x−5).....(ii)

When y=0, then

28=3x−1543=3xx=433

Then the point of intersection with the x-axis, that is point R433,0

Step 4: Calculate the Area PQR,

Area of triangle =12×base×height

Using the figure, Height =7, and

base=433--14=172+312=17512

Therefore, the area A=12×17512×7, then

24A=24×12×17512×7=175×7=1225

Hence, 24A=1225squareunit.


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