Find the transverse common tangents of the circles $x^2 + y^2 - 4x - 10y + 28 = 0$ and $x^2 + y^2 + 4x - 6y + 4 = 0$.

Find the transverse common tangents of the circles – Class 12 Mathematics

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x^2 + y^2 - 4x - 10y + 28 = 0

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 7mImportance★★★★★

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Concept understanding — Angle Between Two Intersecting Circles

The angle between two intersecting circles at a common point $P$ is defined as the angle between their tangents at $P$ -- equivalently (since each tangent is perpendicular to its own radius), the angle between the two radii $C_1P$ and $C_2P$ , where $C_1,C_2$ are the centres.

Let the circles have centres $C_1,C_2$ , radii $r_1,r_2$ , and let $d=C_1C_2$ be the distance between the centres. In triangle $C_1PC_2$ , the sides are $C_1P=r_1$ , $C_2P=r_2$ , and $C_1C_2=d$ , and the included angle at $P$ is the angle $\theta$ between the circles. The cosine rule in this triangle gives

d^2=r_1^2+r_2^2-2r_1r_2\cos\theta\qquad\Longrightarrow\qquad \cos\theta=\frac{d^2-r_1^2-r_2^2}{2r_1r_2}.

This can be rewritten purely in terms of the circles' coefficients. For $S\equiv x^2+y^2+2gx+2fy+c=0$ and $S'\equiv x^2+y^2+2g'x+2f'y+c'=0$ , the centres are $(-g,-f)$ and $(-g',-f')$ , so $d^2=(g-g')^2+(f-f')^2$ , while $r_1^2=g^2+f^2-c$ and $r_2^2=g'^2+f'^2-c'$ . Substituting and expanding:

d^2-r_1^2-r_2^2=(g-g')^2+(f-f')^2-(g^2+f^2-c)-(g'^2+f'^2-c')=c+c'-2gg'-2ff',

since the squared terms $g^2,g'^2,f^2,f'^2$ all cancel. Hence …

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Centres $(2,5),(-2,3)$ with radii $1,3$ ; internal point $(1,\tfrac92)$ ; tangents from it are $x=1$ and $3x+4y-21=0$ .

Circle 1: centre $C_1=(2,5)$ , $r_1=\sqrt{4+25-28}=1$ .

Circle 2: centre $C_2=(-2,3)$ , $r_2=\sqrt{4+9-4}=3$ .

The transverse common tangents meet at the internal centre of similitude $P$ , dividing $C_1C_2$ internally in ratio $r_1:r_2=1:3$ :

$P=\left(\dfrac{r_2\,C_1+r_1\,C_2}{r_1+r_2}\right)=\left(\dfrac{3(2)+1(-2)}{4},\dfrac{3(5)+1(3)}{4}\right)=\left(1,\dfrac92\right).$

Let a tangent through $P$ be $y-\tfrac92=m(x-1)$ , i.e. $mx-y+\left(\tfrac92-m\right)=0$ . Distance from $C_1=(2,5)$ equals $r_1=1$ :

$\dfrac{\left|2m-5+\tfrac92-m\right|}{\sqrt{m^2+1}}=1\Rightarrow \left|m-\tfrac12\right|=\sqrt{m^2+1}.$

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set 2B2 marks
$x^2 + y^2 + 6x - 10y - 135 = 0$
›Reveal solutionSolution

The angle between two intersecting circles is found from $\cos\theta = \dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}$ , where $d$ is the distance between centres.

Circle 1: $x^2+y^2+6x-10y-135=0 \Rightarrow g_1=3, f_1=-5, c_1=-135$

$r_1=\sqrt{9+25+135}=\sqrt{169}=13,\quad C_1=(-3,5)$

Circle 2: $x^2+y^2-4x+14y-116=0 \Rightarrow g_2=-2, f_2=7, c_2=-116$

$r_2=\sqrt{4+49+116}=\sqrt{169}=13,\quad C_2=(2,-7)$

Distance between centres:

$d^2=(-3-2)^2+(5-(-7))^2=25+144=169$

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CBSE 2018Set 2B2 marks
$x^2 + y^2 - 12x - 6y + 41 = 0$
›Reveal solutionSolution

Use the angle-between-two-circles formula $\cos\theta=\dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}$ with the given circles' centres/radii.

Circle 1: $x^2+y^2-12x-6y+41=0$ has centre $(6,3)$ and $r_1^2=36+9-41=4$ , so $r_1=2$ .

Circle 2: $x^2+y^2+kx+6y-59=0$ has centre $(-k/2,-3)$ and $r_2^2=\dfrac{k^2}{4}+9+59=\dfrac{k^2}{4}+68$ .

Distance squared between centres: $d^2=\left(6+\dfrac{k}{2}\right)^2+36$ .

$r_1^2+r_2^2-d^2 = 4+\dfrac{k^2}{4}+68-\left(6+\dfrac{k}{2}\right)^2-36 = -6k$ (the $k^2$ terms and constants cancel).

Using $\cos45^\circ=\dfrac{1}{\sqrt2}$ :

$\dfrac{-6k}{2(2)r_2}=\dfrac{1}{\sqrt2}\Rightarrow -6k\sqrt2=4r_2$ .

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Q.Find the transverse common tangents of the circles x2+y2−4x−10y+28=0x^2 + y^2 - 4x - 10y + 28 = 0x2+y2−4x−10y+28=0 and x2+y2+4x−6y+4=0x^2 + y^2 + 4x - 6y + 4 = 0x2+y2+4x−6y+4=0.

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Q.Find the transverse common tangents of the circles $x^2 + y^2 - 4x - 10y + 28 = 0$ and $x^2 + y^2 + 4x - 6y + 4 = 0$ .

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