# Find the condition for the line px + qy + r = 0 to touch the circle \[{x^2} + {y^2} = {a^2}\]. And also find the coordinates of the point of contact also.

Last updated date: 21st Mar 2023

•

Total views: 306k

•

Views today: 2.84k

Answer

Verified

306k+ views

Hint: Find the equation of tangent to the circle and equate it to the equation of the line to find the point of contact. Then substitute the point in the circle equation to find the condition.

Let us find the slope of the tangent to the circle which is \[\dfrac{{dy}}{{dx}}\].

Given equation of circle is as follows:

\[{x^2} + {y^2} = {a^2}...........(1)\]

Differentiating both sides with respect to x, we get:

\[2x + 2y\dfrac{{dy}}{{dx}} = 0\]

Taking 2x to the other side and simplifying, we get:

\[2y\dfrac{{dy}}{{dx}} = - 2x\]

Now, taking 2y to the other side and finding the slope as follows:

\[\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{y}\]

Let the tangent intersect the circle at the point (h, k), then we have the slope of the tangent as:

\[\dfrac{{dy}}{{dx}} = \dfrac{{ - h}}{k}.........(2)\]

The equation of the line having slope m and passing through a point (a, b) is given by:

\[y - b = m(x - a)\]

Using equation (2) for the equation of tangent in the above equation, we get:

\[y - k = \dfrac{{ - h}}{k}(x - h)\]

Simplifying this expression by multiplying k on both sides, we get:

\[ky - {k^2} = - hx + {h^2}\]

Gathering constant terms on the right-hand side, we have:

\[ky + hx = {h^2} + {k^2}..........(3)\]

The point (h, k) lies on the circle, therefore, using equation (1), we have:

\[{h^2} + {k^2} = {a^2}..........(4)\]

Substituting equation (4) in equation (3), we get:

\[ky + hx = {a^2}\]

\[ky + hx - {a^2} = 0..........(5)\]

The equation of the line is given by,

\[px + qy + r = 0..........(6)\]

From equation (5) and (6), we have:

\[\dfrac{p}{h} = \dfrac{q}{k} = - \dfrac{r}{{{a^2}}}\]

Hence, the value of h and k are as follows:

\[h = - \dfrac{{{a^2}}}{r}p...........(7)\]

\[k = - \dfrac{{{a^2}}}{r}q..........(8)\]

Hence, the point of intersection is \[\left( { - \dfrac{{{a^2}}}{r}p, - \dfrac{{{a^2}}}{r}q} \right)\].

Now, substitute equation (7) and (8) in equation (4).

\[{\left( { - \dfrac{{{a^2}}}{r}p} \right)^2} + {\left( { - \dfrac{{{a^2}}}{r}q} \right)^2} = {a^2}\]

\[\dfrac{{{a^4}}}{{{r^2}}}{p^2} + \dfrac{{{a^4}}}{{{r^2}}}{q^2} = {a^2}\]

\[\dfrac{{{a^4}}}{{{r^2}}}({p^2} + {q^2}) = {a^2}\]

\[{p^2} + {q^2} = \dfrac{{{r^2}}}{{{a^2}}}\]

Hence, the required condition is \[{p^2} + {q^2} = \dfrac{{{r^2}}}{{{a^2}}}\].

Note: You can directly solve the equation without finding the slope of the tangent if you know the formula for the equation of tangent to the circle at a point (h, k), that is, \[hx + ky - {a^2} = 0\].

__Complete step-by-step answer:__Let us find the slope of the tangent to the circle which is \[\dfrac{{dy}}{{dx}}\].

Given equation of circle is as follows:

\[{x^2} + {y^2} = {a^2}...........(1)\]

Differentiating both sides with respect to x, we get:

\[2x + 2y\dfrac{{dy}}{{dx}} = 0\]

Taking 2x to the other side and simplifying, we get:

\[2y\dfrac{{dy}}{{dx}} = - 2x\]

Now, taking 2y to the other side and finding the slope as follows:

\[\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{y}\]

Let the tangent intersect the circle at the point (h, k), then we have the slope of the tangent as:

\[\dfrac{{dy}}{{dx}} = \dfrac{{ - h}}{k}.........(2)\]

The equation of the line having slope m and passing through a point (a, b) is given by:

\[y - b = m(x - a)\]

Using equation (2) for the equation of tangent in the above equation, we get:

\[y - k = \dfrac{{ - h}}{k}(x - h)\]

Simplifying this expression by multiplying k on both sides, we get:

\[ky - {k^2} = - hx + {h^2}\]

Gathering constant terms on the right-hand side, we have:

\[ky + hx = {h^2} + {k^2}..........(3)\]

The point (h, k) lies on the circle, therefore, using equation (1), we have:

\[{h^2} + {k^2} = {a^2}..........(4)\]

Substituting equation (4) in equation (3), we get:

\[ky + hx = {a^2}\]

\[ky + hx - {a^2} = 0..........(5)\]

The equation of the line is given by,

\[px + qy + r = 0..........(6)\]

From equation (5) and (6), we have:

\[\dfrac{p}{h} = \dfrac{q}{k} = - \dfrac{r}{{{a^2}}}\]

Hence, the value of h and k are as follows:

\[h = - \dfrac{{{a^2}}}{r}p...........(7)\]

\[k = - \dfrac{{{a^2}}}{r}q..........(8)\]

Hence, the point of intersection is \[\left( { - \dfrac{{{a^2}}}{r}p, - \dfrac{{{a^2}}}{r}q} \right)\].

Now, substitute equation (7) and (8) in equation (4).

\[{\left( { - \dfrac{{{a^2}}}{r}p} \right)^2} + {\left( { - \dfrac{{{a^2}}}{r}q} \right)^2} = {a^2}\]

\[\dfrac{{{a^4}}}{{{r^2}}}{p^2} + \dfrac{{{a^4}}}{{{r^2}}}{q^2} = {a^2}\]

\[\dfrac{{{a^4}}}{{{r^2}}}({p^2} + {q^2}) = {a^2}\]

\[{p^2} + {q^2} = \dfrac{{{r^2}}}{{{a^2}}}\]

Hence, the required condition is \[{p^2} + {q^2} = \dfrac{{{r^2}}}{{{a^2}}}\].

Note: You can directly solve the equation without finding the slope of the tangent if you know the formula for the equation of tangent to the circle at a point (h, k), that is, \[hx + ky - {a^2} = 0\].

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE