Q23 of 26 Page 1

Using integration, find the area of the triangle formed by the negative x-axis and tangent and normal to the circle at


Equation of circle is given by x2 + y2 = 9


Differentiate with respect to x, we get





The slope of the tangent at (-1, 2√2)



so, the equation of tangent through (-1,2√2) is


-x + 2√2y = 9


⇒ x - 2√2y + 9 = 0


…(i)


It cuts x – axis at (-9,0)


Now, equation of normal to the circle is


(y - 2√2) = slope of normal (x – (-1))


[We know, the slope of the normal × slope of tangent = -1


So, the slope of normal = ]


now, equation of normal is


(y - 2√2) = -2√2 (x + 1)


⇒ y - 2√2 = -2√2x - 2√2


⇒ y = -2√2x …(ii)


Area of ΔOPB












= 9√2 sq. unit

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