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Question: How do you find an equation for the line tangent to the circle \( {x^2} + {y^2} = 25 \) at the point...

How do you find an equation for the line tangent to the circle x2+y2=25{x^2} + {y^2} = 25 at the point (3,−4)(3, - 4) ?

Explanation

Solution

Hint : In this question, we are given the equation of a circle and we have to find the equation of the tangent at the point (3,−4)(3, - 4) , so to solve this question, we must know what a tangent is and how can we find the equation of a line. A line that is perpendicular to the radius of the circle at a point on the circle is said to be the tangent at that point. Using this relation, we can find the slope of the tangent. As the tangent passes through (3,−4)(3, - 4) so we can find the equation of the tangent using the passing point and the slope.

Complete step-by-step answer :
The slope of the radius is equal to,
m1=0−(−4)0−3=4−3{m_1} = \dfrac{{0 - ( - 4)}}{{0 - 3}} = \dfrac{4}{{ - 3}}
We know that the product of the slopes of two perpendicular lines is equal to -1, so the slope of the tangent will be negative of the reciprocal of the slope of the radius, that is, the slope of the tangent is –
m=−1m1=−14−3=−1×−34=34m = \dfrac{{ - 1}}{{{m_1}}} = \dfrac{{ - 1}}{{\dfrac{4}{{ - 3}}}} = - 1 \times \dfrac{{ - 3}}{4} = \dfrac{3}{4}
Now the equation of a line passing through point (x1,y1)({x_1},{y_1}) and having a slope m is given as –
y−y1=m(x−x1)y - {y_1} = m(x - {x_1})
So, the equation of tangent is –
y−(−4)=34(x−3) ⇒4(y+4)=3(x−3) ⇒4y+16=3x−9 ⇒3x−4y−25=0   y - ( - 4) = \dfrac{3}{4}(x - 3) \\\ \Rightarrow 4(y + 4) = 3(x - 3) \\\ \Rightarrow 4y + 16 = 3x - 9 \\\ \Rightarrow 3x - 4y - 25 = 0 \;
Hence, the equation for the line tangent to the circle x2+y2=25{x^2} + {y^2} = 25 at the point (3,−4)(3, - 4) is given as 3x−4y−25=03x - 4y - 25 = 0
So, the correct answer is “ 3x−4y−25=03x - 4y - 25 = 0 ”.

Note : Comparing the given equation of the circle with the standard equation of the circle, that is, (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
We get that the coordinates of the centre of the circle is (0,0)(0,0) and its radius is 5 units.
We know two passing points of the radius of the circle, so its slope is given as –
m1=ΔyΔx=0−(−4)0−3=4−3{m_1} = \dfrac{{\Delta y}}{{\Delta x}} = \dfrac{{0 - ( - 4)}}{{0 - 3}} = \dfrac{4}{{ - 3}}
This is how we find the slope of a line, and we can solve similar questions using this approach.